Methods of Solving First Order, First Degree Differential Equation

Q: What are the methods of solving first order first degree differential equations?

The standard methods are variables separable, homogeneous equations solved with y = vx, exact equations, and linear equations solved with an integrating factor. Substitutions such as t = ax + by + c, a shift of origin, polar coordinates and Bernoulli's z = y^(1-n) reduce other equations to one of these four types.

Q: How do you know if a differential equation is homogeneous?

Write it as dy/dx = (f(x, y))/(g(x, y)). It is homogeneous if f and g are homogeneous functions of the same degree, which means replacing x by kx and y by ky multiplies both by the same power of k. Then dy/dx depends only on y/x and the substitution y = vx separates the variables.

Q: What is an integrating factor in a linear differential equation?

For dy/dx + Py = Q, the integrating factor is e raised to the integral of P dx. Multiplying by it turns the left side into the derivative of y times the factor, so y times IF equals the integral of Q times IF plus C. No constant is added while finding the factor itself.

Q: How do you check whether a differential equation is exact?

Write the equation as M dx + N dy = 0. It is exact when the partial derivative of M with respect to y equals the partial derivative of N with respect to x. Then M dx + N dy is the total differential of some F(x, y), and the solution is F(x, y) = c.

Q: What is Bernoulli's differential equation and how is it solved?

Bernoulli's equation is dy/dx + Py = Qy^n with n not equal to 0 or 1, where P and Q depend on x only. Divide by y^n and substitute z = y^(1-n). The equation becomes dz/dx + (1 - n)Pz = (1 - n)Q, which is linear in z and is solved with the integrating factor.

Q: When should x be taken as the dependent variable in a differential equation?

When the equation is not linear in y but x appears only to the first power, rewrite it for dx/dy. For example, (x + 2y³)dy/dx = y becomes dx/dy - x/y = 2y², a linear equation in x with integrating factor e raised to the integral of P1 dy, here 1/y.

Q: What is the singular solution of Clairaut's equation?

Clairaut's equation y = px + f(p), with p = dy/dx, has the general solution y = cx + f(c), a family of straight lines. Eliminating p between the equation and x + f'(p) = 0 gives the singular solution, which contains no constant and is the envelope that touches every one of those lines.

Q: Which differential equation methods are asked in JEE Main?

The JEE Main syllabus names solution by separation of variables, homogeneous equations and linear equations of the type dy/dx + P(x)y = Q(x), so these three appear most often, usually with an initial condition. Exact equations, Bernoulli's equation, substitutions and orthogonal trajectories are common in JEE Advanced level problems.

MathsDifferential EquationsFor JEE aspirants

Methods of solving a first order, first degree differential equation turn $\frac{d y}{d x} = f (x, y)$ into something you can integrate. The four standard methods are variables separable, homogeneous ( $y = v x$ ), exact, and linear (integrating factor). Substitutions such as $t = a x + b y + c$ , a shift of origin, polar coordinates and Bernoulli's $z = y^{1 - n}$ reduce other forms to these. The page ends with Clairaut's equation, orthogonal trajectories and geometric applications of the first order first degree differential equation. Separable, homogeneous and linear equations are named in the JEE Main syllabus.

On this page1Choose a method2Separable3Homogeneous4Exact5Linear and Bernoulli6Clairaut7Orthogonal trajectories8Geometry

Key Formulas - Quick Reference

★ Must learnSeparable: $f (x) d x = ϕ (y) d y \Rightarrow \int f (x) d x = \int ϕ (y) d y + c$ .
$\frac{d y}{d x} = f (a x + b y + c)$ : put $t = a x + b y + c$ , so $\frac{d t}{d x} = a + b f (t)$ .
★ Must learnHomogeneous $\frac{d y}{d x} = F (\frac{y}{x})$ : put $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
$\frac{d y}{d x} = \frac{a x + b y + c}{A x + B y + C}$ : shift $x = X + h$ , $y = Y + k$ to the meeting point of the two lines.
★ Must learnExact $M d x + N d y = 0$ when $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ .
★ Must learnLinear $\frac{d y}{d x} + P y = Q$ : IF $= e^{\int P d x}$ , and $y \cdot IF = \int Q \cdot IF d x + C$ .
★ Must learnBernoulli $\frac{d y}{d x} + P y = Q y^{n}$ : put $z = y^{1 - n}$ ; the equation becomes linear in $z$ .
Clairaut $y = p x + f (p)$ : general $y = c x + f (c)$ ; singular from $x + f^{'} (p) = 0$ .
Orthogonal trajectory: replace $\frac{d y}{d x}$ by $- \frac{d x}{d y}$ in the family's equation.
Polar: $x d x + y d y = r d r$ , $x d y - y d x = r^{2} d θ$ .

1. First Order, First Degree Differential Equations

A differential equation of first order and first degree has the form

\frac{d y}{d x} + f (x, y) = 0

, which can also be written as

M d x + N d y = 0

, where

M

and

N

are functions of

x

and

y

No single formula solves every such equation. Instead, recognise which standard form it fits. Figure 1 gives the order of tests that works for almost every exam question; the sections below take each exit in turn.

Figure 1: Run the four tests in this order. Most exam questions stop at the first or second diamond; the last box turns a stubborn equation into one of the four standard types.

2. Variables Separable

2.1 The method

If the differential equation can be put in the form

f (x) d x = ϕ (y) d y

, the variables are separable, and the solution is obtained by integrating each side separately:

\int f (x) d x = \int ϕ (y) d y + c,

where

c

is an arbitrary constant.

Typical sign: $\frac{d y}{d x} = g (x) h (y)$ , a product of a function of $x$ and a function of $y$ . Divide by $h (y)$ , multiply by $d x$ , integrate. One constant is enough: constants from the two sides merge into one (Solved Examples 1 to 3).

Dividing by

h (y)

assumes

h (y) \neq = 0

. Any constant root of

h (y) = 0

is itself a solution. In Figure 2, dividing by

y

hides the solution

y = 0

Figure 2: Separating

\frac{d y}{y} = x d x

and integrating gives

y = C e^{x^{2} /2}

. Dividing by

y

assumed

y \neq = 0

; the dashed line

y = 0

is also a solution (

C = 0

2.2 Polar coordinate substitution

Sometimes a change to polar coordinates makes the variables separate. Remember these differentials:

(a) $x = r cos θ$ , $y = r sin θ$ : (i) $x d x + y d y = r d r$ (ii) $d x^{2} + d y^{2} = d r^{2} + r^{2} d θ^{2}$ (iii) $x d y - y d x = r^{2} d θ$
(b) $x = r sec θ$ , $y = r tan θ$ : (i) $x d x - y d y = r d r$ (ii) $x d y - y d x = r^{2} sec θ d θ$

Where they come from: $x^{2} + y^{2} = r^{2}$ gives (a)(i) on differentiating, and $\frac{y}{x} = tan θ$ gives $\frac{x d y - y d x}{x ^{2}} = sec^{2} θ d θ$ , which is (a)(iii). For (b), $x^{2} - y^{2} = r^{2}$ and $\frac{y}{x} = sin θ$ .

Figure 3: When an equation contains

x d x + y d y

and

x d y - y d x

, polar coordinates replace them by

r d r

and

r^{2} d θ

, and the variables separate.

In Solved Example 4, $x d x + y d y = x (x d y - y d x)$ becomes $\frac{d r}{r ^{2}} = cos θ d θ$ . Its solutions turn out to be conics with a common focus at the origin (Figure 4).

Figure 4:

(y + 1)^{2} = k (x^{2} + y^{2})

says distance from

O

= \frac{1}{k} \times

distance from the line

y = - 1

. So every solution is a conic with focus

O

, directrix

y = - 1

and eccentricity

e = \frac{1}{k}

2.3 Equations reducible to variables separable

If a proper substitution reduces an equation to separable form, it is called reducible to the variables separable type. The standard case is

\frac{d y}{d x} = f (a x + b y + c), b \neq = 0.

Put $a x + b y + c = t$ .
Differentiate: $a + b \frac{d y}{d x} = \frac{d t}{d x}$ , so $\frac{d t}{d x} = a + b f (t)$ .
Separate: $\frac{d t}{a + b f ( t )} = d x$ , integrate, and put back $t = a x + b y + c$ .

Solved Examples 5 and 6 use $t = 4 x + y + 1$ and $t = x + y$ .

Exam Trick

A repeated bracket is a substitution waiting to happen. If $x + y$ , $x - y$ or $a x + b y$ appears inside $sin$ , a square or an exponent, make it the new variable. For $\int \frac{d t}{1 + sin t}$ multiply top and bottom by $1 - sin t$ to get $\int (sec^{2} t - sec t tan t) d t$ .

Key idea

Separable: split into

f (x) d x = ϕ (y) d y

and integrate once. If a combination such as

a x + b y + c

x^{2} + y^{2}

keeps appearing, substitute first.

3. Homogeneous Differential Equations

3.1 Homogeneous functions

A function

f (x, y)

is homogeneous of degree $n$ if

f (λ x, λ y) = λ^{n} f (x, y)

for every

λ > 0

. Then

f (x, y) = x^{n} ϕ (\frac{y}{x})

Examples: $x^{2} + x y - 3 y^{2}$ (degree 2), $x^{2} + y^{2}$ (degree 1), $\frac{x}{y} + sin \frac{y}{x}$ (degree 0). Not homogeneous: $x^{2} + y$ , $x + cos y$ .

3.2 Solving by $y = v x$

An equation

\frac{d y}{d x} = \frac{f ( x , y )}{g ( x , y )}

, where

f

and

g

are homogeneous functions of the same degree, is called a homogeneous differential equation. The right side is then a function of

\frac{y}{x}

only, and it is solved by putting $y = v x$ .

Put $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
The equation becomes $v + x \frac{d v}{d x} = F (v)$ , i.e. $x \frac{d v}{d x} = F (v) - v$ .
Separate: $\int \frac{d v}{F ( v ) - v} = \int \frac{d x}{x}$ .
Replace $v$ by $\frac{y}{x}$ .

If the equation is naturally $\frac{d x}{d y} = G (\frac{x}{y})$ , put $x = v y$ instead (Solved Example 28).

Put

y = v x

Form $\frac{d y}{d x} = F (\frac{y}{x})$ .
$\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
Separate $v$ and $x$ .

Put

x = v y

Form $\frac{d x}{d y} = G (\frac{x}{y})$ .
$\frac{d x}{d y} = v + y \frac{d v}{d y}$ .
Separate $v$ and $y$ .

Figure 5: In a homogeneous equation the slope is a function of

v = \frac{y}{x}

only, so it is constant along each ray

y = v x

. Solved Example 8: the circles

x^{2} + y^{2} = c x

, and the one through

(1, 1)

3.3 Equations reducible to homogeneous form

Equations of the form

\frac{d y}{d x} = \frac{a x + b y + c}{A x + B y + C} (1)

can be made homogeneous in new variables $X$ , $Y$ by substituting $x = X + h$ and $y = Y + k$ , where $h$ and $k$ are constants. Since $\frac{d y}{d x} = \frac{d Y}{d X}$ ,

\frac{d Y}{d X} = \frac{a X + bY + ( ah + bk + c )}{A X + B Y + ( A h + B k + C )} (2)

Choose $h$ , $k$ so that $ah + bk + c = 0$ and $A h + B k + C = 0$ , i.e. $(h, k)$ is the meeting point of the two lines. Then $\frac{d Y}{d X} = \frac{a X + bY}{A X + B Y}$ , which is solved by $Y = v X$ (Solved Example 9). This needs the lines to meet: $\frac{a}{A} \neq = \frac{b}{B}$ .

Figure 6: Solved Example 9. Numerator and denominator are two straight lines; moving the origin to their meeting point

(1, 2)

removes the constants

- 5

and

- 4

Special cases of (1):

Case 1: if $\frac{a}{A} = \frac{b}{B}$ (parallel lines), the substitution $a x + b y = v$ makes the variables separable (Solved Example 10).
Case 2: if $b + A = 0$ , cross-multiplying turns (1) into an exact equation (Solved Example 11).
Case 3: an equation of the form $y f (x y) d x + x g (x y) d y = 0$ (not homogeneous, but a close relative) separates with $x y = v$ : it becomes $\frac{d x}{x} = \frac{g ( v ) d v}{v [ g ( v ) - f ( v )]}$ .

Figure 7: Check the two shortcuts before shifting the origin. Solved Example 10 takes the first exit, Example 11 the second and Example 9 the third.

Quick Recall: tap to check

\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{x y}

homogeneous?

Yes: numerator and denominator both have degree 2.

After

y = v x

, what is

\frac{d y}{d x}

v + x \frac{d v}{d x}

(not

v + \frac{d v}{d x}

For

\frac{d y}{d x} = \frac{x + y + 1}{2 x + 2 y + 3}

, which route?

Case 1:

\frac{1}{2} = \frac{1}{2}

, so put

u = x + y

For

\frac{d y}{d x} = \frac{x + 2 y - 5}{2 x + y - 4}

, where do the new axes go?

At the meeting point

(1, 2)

x + 2 y = 5

and

2 x + y = 4

Key idea

Homogeneous means the slope depends only on

\frac{y}{x}

: put

y = v x

. Linear fractions become homogeneous after moving the origin to the meeting point of the two lines.

4. Exact Differential Equations

The equation

M + N \frac{d y}{d x} = 0

, i.e.

M d x + N d y = 0

, where

M

and

N

are functions of

x

and

y

, is exact if it can be derived by direct differentiation (without any later multiplication, elimination and so on) of an equation of the form

f (x, y) = c

Example: $y^{2} d y + x d x + \frac{d x}{x} = 0$ is exact, being $d (\frac{y ^{3}}{3} + \frac{x ^{2}}{2} + ln x) = 0$ .

Test:

M d x + N d y = 0

is exact if and only if

\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}

(for

M

N

with continuous partial derivatives in a rectangle). The condition is necessary and also sufficient.

Figure 8: Differentiate each coefficient with respect to the other variable. Equal results mean

M d x + N d y

is an exact differential

d F

(Solved Example 13).

Solving an exact equation. Either group the terms into standard exact differentials, or use $\int M d x$ (treating $y$ as constant) $+ \int$ (terms of $N$ free of $x$ ) $d y = c$ . These exact differentials should be remembered:

Expression	Exact differential
(a) $x d y + y d x$	$d (x y)$
(b) $\frac{x d y - y d x}{x ^{2}}$	$d (\frac{y}{x})$
$\frac{y d x - x d y}{y ^{2}}$	$d (\frac{x}{y})$
(c) $2 (x d x + y d y)$	$d (x^{2} + y^{2})$
(d) $\frac{x d y - y d x}{x y}$	$d (ln \frac{y}{x})$
(e) $\frac{x d y - y d x}{x ^{2} + y ^{2}}$	$d (tan^{- 1} \frac{y}{x})$
(f) $\frac{x d y + y d x}{x y}$	$d (ln x y)$
(g) $\frac{x d y + y d x}{x ^{2} y ^{2}}$	$d (- \frac{1}{x y})$

Exam Trick

Look for the pairs $x d y + y d x$ and $x d y - y d x$ . Once you see one, divide the whole equation by whatever makes it a row of the table: $x y$ , $x^{2}$ , $y^{2}$ or $x^{2} + y^{2}$ . For $x d y + y d x + x y e^{y} d y = 0$ , dividing by $x y$ gives $d (ln x y) + d (e^{y}) = 0$ (Practice Question 13).

JEE Advanced

Integrating factor for a non-exact equation. If $\frac{1}{N} (\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}) = f (x)$ depends on $x$ alone, multiplying by $e^{\int f (x) d x}$ makes $M d x + N d y = 0$ exact. Similarly, if $\frac{1}{M} (\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}) = g (y)$ , use $e^{\int g (y) d y}$ . The linear equation is the special case: $(P y - Q) d x + d y = 0$ gives $f (x) = P$ and the factor $e^{\int P d x}$ .

5. Linear Differential Equations

5.1 What makes an equation linear

The dependent variable and its derivatives occur in the first degree only and are not multiplied together.
All the derivatives are in polynomial form.
The order may be more than one.

The $m$ th order linear differential equation has the form

P_{0} (x) \frac{d ^{m} y}{d x ^{m}} + P_{1} (x) \frac{d ^{m - 1} y}{d x ^{m - 1}} + \dots + P_{m - 1} (x) \frac{d y}{d x} + P_{m} (x) y = ϕ (x),

where $P_{0} (x), P_{1} (x), \dots, P_{m} (x)$ are the coefficients. Note: $\frac{d y}{d x} + y^{2} sin x = ln x$ is not linear, because of $y^{2}$ .

5.2 First order linear equation and the integrating factor

\frac{d y}{d x} + P y = Q

, where

P

and

Q

are functions of

x

only, is linear in $y$ . An integrating factor (IF) is an expression which, when multiplied into a differential equation, makes it exact. Here $IF = e^{\int P d x}$ (no constant of integration is needed).

Multiplying by the IF:

\frac{d y}{d x} e^{\int P d x} + P y e^{\int P d x} = Q e^{\int P d x} \Rightarrow \frac{d}{d x} (y e^{\int P d x}) = Q e^{\int P d x}

y e^{\int P d x} = \int Q e^{\int P d x} d x + C

Before finding the IF, divide so that the coefficient of $\frac{d y}{d x}$ is 1 (Solved Example 15).

Figure 9: With

P = 2 x

the integrating factor is

e^{x^{2}}

, and

\frac{d}{d x} (y e^{x^{2}}) = 2 x e^{x^{2}}

gives

y = 1 + C e^{- x^{2}}

. Every member is the particular solution

y = 1

plus

C

times

e^{- \int P d x}

$P$	$\int P d x$	IF
$\frac{k}{x}$	$k ln x$	$x^{k}$
$tan x$	$ln sec x$	$sec x$
$cot x$	$ln sin x$	$sin x$
$\frac{1}{x ln x}$	$ln (ln x)$	$ln x$
$\frac{3 x ^{2}}{1 + x ^{3}}$	$ln (1 + x^{3})$	$1 + x^{3}$
$2 x$	$x^{2}$	$e^{x^{2}}$

5.3 Linear in $x$

Sometimes the equation becomes linear if $x$ is taken as the dependent variable and $y$ as the independent one: $\frac{d x}{d y} + P_{1} x = Q_{1}$ , where $P_{1}$ and $Q_{1}$ are functions of $y$ . The IF is then $e^{\int P_{1} d y}$ and $x e^{\int P_{1} d y} = \int Q_{1} e^{\int P_{1} d y} d y + C$ (Solved Examples 16 and 29).

Exam Trick

Not linear in $y$ ? Flip it. If $y$ appears as $y^{3}$ , $e^{y}$ or $ln y$ but $x$ appears only to the first power, write the equation for $\frac{d x}{d y}$ . For $(x + 2 y^{3}) \frac{d y}{d x} = y$ : $\frac{d x}{d y} - \frac{x}{y} = 2 y^{2}$ , linear in $x$ .

5.4 Reducible to linear by a change of variable

An equation of the form $f^{'} (y) \frac{d y}{d x} + P f (y) = Q$ becomes linear in $z$ with $z = f (y)$ , since $\frac{d z}{d x} = f^{'} (y) \frac{d y}{d x}$ . Example: $y sin x \frac{d y}{d x} = cos x (sin x - y^{2})$ with $z = y^{2}$ (Solved Example 17).

5.5 Bernoulli's equation

\frac{d y}{d x} + P y = Q y^{n}

n \neq = 0

and

n \neq = 1

, where

P

and

Q

are functions of

x

, is Bernoulli's equation. Divide by

y^{n}

and put

z = y^{1 - n}

(i.e.

y^{- n + 1} = z

): it becomes linear in $z$ ,

\frac{d z}{d x} + (1 - n) P z = (1 - n) Q,

and is solved with the IF of Section 5.2.

Example of the form: $2 sin x \frac{d y}{d x} - y cos x = x y^{3} e^{x}$ ( $n = 3$ , put $z = y^{- 2}$ ). Solved Example 18 has $n = 2$ .

Linear

$\frac{d y}{d x} + P y = Q$ .
$y$ to power 1 only.
Multiply by $e^{\int P d x}$ directly.

Bernoulli

$\frac{d y}{d x} + P y = Q y^{n}$ .
Extra $y^{n}$ on the right.
First $z = y^{1 - n}$ , then the IF.

Quick Recall: tap to check

IF of

\frac{d y}{d x} - \frac{y}{x} = x^{2}

e^{- l n x} = \frac{1}{x}

Which substitution linearises

\frac{d y}{d x} + \frac{y}{x} = y^{3}

z = y^{- 2}

(here

n = 3

\frac{d y}{d x} + x y = sin x

linear? Is

\frac{d y}{d x} + x sin y = 0

The first is linear in

y

; the second is not (it is separable).

Key idea

Linear: make the coefficient of

\frac{d y}{d x}

equal to 1, multiply by

e^{\int P d x}

, and the left side collapses to

\frac{d}{d x} (y \cdot IF)

. Bernoulli and change-of-variable equations are linear in disguise.

6. Clairaut's Equation

The equation

y = m x + f (m)

, where

m = \frac{d y}{d x}

(often written

p

), is Clairaut's equation. It is of first order, though usually not of first degree.

Differentiate with respect to $x$ :

\frac{d y}{d x} = m + x \frac{d m}{d x} + f^{'} (m) \frac{d m}{d x} \Rightarrow \frac{d m}{d x} [x + f^{'} (m)] = 0

So either $\frac{d m}{d x} = 0$ , i.e. $m = c$ ... (2), or $x + f^{'} (m) = 0$ ... (3).

Eliminating $m$ between the equation and (2) gives the general solution $y = c x + f (c)$ : just replace $m$ by $c$ . It is a family of straight lines.
Eliminating $m$ between the equation and (3) gives a solution with no arbitrary constant that is not a particular solution: the singular solution, the envelope of those lines.

Figure 10: Solved Example 19. Replacing

p

c

gives the straight lines (general solution); eliminating

p

with

x + 1 - 3 p^{2} = 0

gives their envelope

27 y^{2} = 4 (x + 1)^{3}

Some equations become Clairaut's after rearranging. In Practice Question 20, $sin p x cos y = cos p x sin y + p$ is $sin (p x - y) = p$ , i.e. $y = p x - sin^{- 1} p$ .

7. Orthogonal Trajectories

An orthogonal trajectory of a given system of curves is a curve that cuts every member of the family at right angles.

Let $f (x, y, c) = 0$ be the given family, $c$ an arbitrary constant.
Differentiate with respect to $x$ and eliminate $c$ .
Replace $\frac{d y}{d x}$ by $- \frac{d x}{d y}$ in the equation from step 2.
Solve the new differential equation: its solution is the family of orthogonal trajectories.

The swap in step 3 works because perpendicular slopes multiply to $- 1$ . Lines through the origin have circles centred at the origin as orthogonal trajectories (Solved Example 20); the parabolas $y^{2} = 4 a x$ have ellipses (Figure 11).

Figure 11: The family

y^{2} = 4 a x

gives

y = 2 x \frac{d y}{d x}

; replacing

\frac{d y}{d x}

- \frac{d x}{d y}

and integrating gives the ellipses

2 x^{2} + y^{2} = c

, which meet each parabola at

9 0^{\circ}

Key idea

Orthogonal trajectories: form the family's equation, eliminate the constant, then replace

\frac{d y}{d x}

- \frac{d x}{d y}

and solve.

8. Geometrical Applications

Many problems describe a curve by a property of its tangent or normal. Turn the property into a differential equation, then solve it with the methods above. With $y$ the ordinate of $P$ and $m = \frac{d y}{d x}$ the slope of the tangent at $P$ :

Quantity	Formula
(i) Length of tangent $L_{T} = P T$	$\frac{y 1 + m ^{2}}{m}$
(ii) Length of normal $L_{N} = P G$	$y 1 + m^{2}$
(iii) Length of subtangent $L_{S T} = T N$	$\frac{y}{m}$
(iv) Length of subnormal $L_{S N} = N G$	$∣ m y ∣$
Intercepts of the tangent on the axes	$x - \frac{y}{m}$ and $y - m x$
Perpendicular from $O$ to the tangent	$\frac{∣ y - m x ∣}{1 + m ^{2}}$

Figure 12: With slope

m = \frac{d y}{d x}

P

T N = \frac{y}{m}

N G = ∣ m y ∣

P T = \frac{y 1 + m ^{2}}{m}

P G = ∣ y ∣ 1 + m^{2}

. Solved Example 22 uses

P G = O P

Quick Recall: tap to check

Subnormal of

y^{2} = 4 a x

at any point?

∣ m y ∣ = y \cdot \frac{2 a}{y} = 2 a

, a constant.

What replaces

\frac{d y}{d x}

for orthogonal trajectories?

- \frac{d x}{d y}

General solution of

y = p x + \frac{2}{p}

y = c x + \frac{2}{c}

; the singular one is

y^{2} = 8 x

Figure 13: Five standard forms and their key moves, plus the applications that use them.

9. Solved Examples

9.1 Variables separable and reducible forms

Solved Example 1

Solve the differential equation

(1 + x) y d x = (y - 1) x d y

Solution:

Separate: $(\frac{1 + x}{x}) d x = (\frac{y - 1}{y}) d y$ .
Integrate: $\int (\frac{1}{x} + 1) d x = \int (1 - \frac{1}{y}) d y$ , so $ln x + x = y - ln y + c$ .
Rearrange: $ln y + ln x = y - x + c$ .

Answer: $x y = C e^{y - x}$

Solved Example 2

Solve

\frac{d y}{d x} = (e^{x} + 1) (1 + y^{2})

Solution:

Separate: $\frac{d y}{1 + y ^{2}} = (e^{x} + 1) d x$ .
Integrate both sides.

Answer: $tan^{- 1} y = e^{x} + x + c$

Solved Example 3

Solve

y - x \frac{d y}{d x} = a (y^{2} + \frac{d y}{d x})

Solution:

Collect: $y - a y^{2} = (x + a) \frac{d y}{d x}$ , so $\frac{d x}{x + a} = \frac{d y}{y ( 1 - a y )}$ .
Partial fractions: $\frac{1}{y ( 1 - a y )} = \frac{1}{y} + \frac{a}{1 - a y}$ .
Integrate: $ln (x + a) = ln y - ln (1 - a y) + ln c = ln \frac{cy}{1 - a y}$ .

Answer: $cy = (x + a) (1 - a y)$ , where $c$ is an arbitrary constant

Solved Example 4

Solve the differential equation

x d x + y d y = x (x d y - y d x)

Solution:

Take $x = r cos θ$ , $y = r sin θ$ . Then $x^{2} + y^{2} = r^{2}$ gives $x d x + y d y = r d r$ ... (i).
$\frac{y}{x} = tan θ$ gives $\frac{x d y - y d x}{x ^{2}} = sec^{2} θ d θ$ , so $x d y - y d x = r^{2} d θ$ ... (ii).
Using (i) and (ii): $r d r = r cos θ \cdot r^{2} d θ$ , so $\frac{d r}{r ^{2}} = cos θ d θ$ .
Integrate: $- \frac{1}{r} = sin θ + λ$ , i.e. $- \frac{1}{x ^{2} + y ^{2}} = \frac{y}{x ^{2} + y ^{2}} + λ$ .
With $c = - λ$ : $\frac{y + 1}{x ^{2} + y ^{2}} = c$ , so $(y + 1)^{2} = c^{2} (x^{2} + y^{2})$ .

Answer: $(y + 1)^{2} = k (x^{2} + y^{2})$ , where $k = c^{2}$ (conics with focus at the origin, Figure 4)

Solved Example 5

Solve

\frac{d y}{d x} = (4 x + y + 1)^{2}

Solution:

Put $4 x + y + 1 = t$ : $4 + \frac{d y}{d x} = \frac{d t}{d x}$ , so $\frac{d y}{d x} = \frac{d t}{d x} - 4$ .
The equation becomes $\frac{d t}{d x} - 4 = t^{2}$ , i.e. $\frac{d t}{t ^{2} + 4} = d x$ (variables separated).
Integrate: $\frac{1}{2} tan^{- 1} \frac{t}{2} = x + c$ .

Answer: $\frac{1}{2} tan^{- 1} (\frac{4 x + y + 1}{2}) = x + c$

Solved Example 6

Solve

sin^{- 1} (\frac{d y}{d x}) = x + y

Solution:

$\frac{d y}{d x} = sin (x + y)$ . Put $x + y = t$ : $\frac{d y}{d x} = \frac{d t}{d x} - 1$ .
$\frac{d t}{d x} - 1 = sin t$ , so $\frac{d t}{1 + sin t} = d x$ .
$\int \frac{1 - sin t}{cos ^{2} t} d t = \int (sec^{2} t - sec t tan t) d t = tan t - sec t = x + c$ .
$- \frac{1 - sin t}{cos t} = x + c$ , so $sin t - 1 = x cos t + c cos t$ .

Answer: $sin (x + y) = x cos (x + y) + c cos (x + y) + 1$

9.2 Homogeneous and reducible to homogeneous

Solved Example 7

Solve

\frac{2 y}{x} + [(\frac{y}{x})^{2} - 1] \frac{d y}{d x} = 0

Solution:

Put $y = v x$ : $2 v + (v^{2} - 1) (v + x \frac{d v}{d x}) = 0$ , so $v + x \frac{d v}{d x} = - \frac{2 v}{v ^{2} - 1}$ .
$x \frac{d v}{d x} = - \frac{v ( 1 + v ^{2} )}{v ^{2} - 1}$ , so $\int \frac{v ^{2} - 1}{v ( 1 + v ^{2} )} d v = - \int \frac{d x}{x}$ .
Partial fractions: $\int (\frac{2 v}{1 + v ^{2}} - \frac{1}{v}) d v = - ln x + c$ , so $ln (1 + v^{2}) - ln v + ln x = c$ .
$ln \frac{1 + v ^{2}}{v} \cdot x = c$ , i.e. $ln \frac{x ^{2} + y ^{2}}{y} = c$ .

Answer: $x^{2} + y^{2} = c^{'} y$ , where $c^{'} = e^{c}$

Solved Example 8

Solve

(x^{2} - y^{2}) d x + 2 x y d y = 0

, given that

y = 1

when

x = 1

Solution:

$\frac{d y}{d x} = - \frac{x ^{2} - y ^{2}}{2 x y}$ . Put $y = v x$ , $\frac{d y}{d x} = v + x \frac{d v}{d x}$ .
$v + x \frac{d v}{d x} = - \frac{1 - v ^{2}}{2 v}$ , so $x \frac{d v}{d x} = - \frac{1 + v ^{2}}{2 v}$ .
$\int \frac{2 v}{1 + v ^{2}} d v = - \int \frac{d x}{x}$ : $ln (1 + v^{2}) = - ln x + c$ .
At $x = 1$ , $y = 1$ : $v = 1$ , so $c = ln 2$ . Then $ln [(1 + \frac{y ^{2}}{x ^{2}}) x] = ln 2$ .

Answer: $x^{2} + y^{2} = 2 x$ (Figure 5)

Solved Example 9

Solve the differential equation

\frac{d y}{d x} = \frac{x + 2 y - 5}{2 x + y - 4}

Solution:

Let $x = X + h$ , $y = Y + k$ ; then $\frac{d y}{d x} = \frac{d Y}{d X}$ and $\frac{d Y}{d X} = \frac{X + 2 Y + ( h + 2 k - 5 )}{2 X + Y + ( 2 h + k - 4 )}$ .
Choose $h + 2 k - 5 = 0$ and $2 h + k - 4 = 0$ : $h = 1$ , $k = 2$ (Figure 6). So $\frac{d Y}{d X} = \frac{X + 2 Y}{2 X + Y}$ , homogeneous.
$Y = v X$ : $X \frac{d v}{d X} = \frac{1 + 2 v}{2 + v} - v = \frac{1 - v ^{2}}{2 + v}$ , so $\int \frac{2 + v}{1 - v ^{2}} d v = \int \frac{d X}{X}$ .
$\int [\frac{1}{2 ( v + 1 )} + \frac{3}{2 ( 1 - v )}] d v = ln X + c$ : $\frac{1}{2} ln (v + 1) - \frac{3}{2} ln (1 - v) = ln X + c$ .
$ln \frac{v + 1}{( 1 - v ) ^{3}} = ln X^{2} + 2 c$ , which gives $X + Y = c^{'} (X - Y)^{3}$ .
Put back $X = x - 1$ , $Y = y - 2$ : $x - 1 + y - 2 = c^{'} (x - 1 - y + 2)^{3}$ .

Answer: $x + y - 3 = c^{'} (x - y + 1)^{3}$

Solved Example 10

Solve

\frac{d y}{d x} = \frac{2 x + 3 y - 1}{4 x + 6 y - 5}

Solution:

Here $\frac{2}{4} = \frac{3}{6}$ (Case 1). Put $u = 2 x + 3 y$ : $\frac{d u}{d x} = 2 + 3 \frac{d y}{d x}$ .
$\frac{1}{3} (\frac{d u}{d x} - 2) = \frac{u - 1}{2 u - 5}$ , so $\frac{d u}{d x} = \frac{3 u - 3 + 4 u - 10}{2 u - 5} = \frac{7 u - 13}{2 u - 5}$ .
$\int \frac{2 u - 5}{7 u - 13} d u = \int d x$ , and $\frac{2 u - 5}{7 u - 13} = \frac{2}{7} - \frac{9}{7} \cdot \frac{1}{7 u - 13}$ .
$\frac{2}{7} u - \frac{9}{49} ln (7 u - 13) = x + c$ . Multiply by 7 and put $u = 2 x + 3 y$ :
$4 x + 6 y - \frac{9}{7} ln (14 x + 21 y - 13) = 7 x + 7 c$

Answer: $- 3 x + 6 y - \frac{9}{7} ln (14 x + 21 y - 13) = c^{'}$

Solved Example 11

Solve

\frac{d y}{d x} = \frac{x - 2 y + 5}{2 x + y - 1}

Solution:

Here $b + A = - 2 + 2 = 0$ (Case 2). Cross-multiplying: $2 x d y + y d y - d y = x d x - 2 y d x + 5 d x$ .
Group: $2 (x d y + y d x) + y d y - d y = x d x + 5 d x$ , i.e. $2 d (x y) + y d y - d y = x d x + 5 d x$ .
Integrate: $2 x y + \frac{y ^{2}}{2} - y = \frac{x ^{2}}{2} + 5 x + c$ .

Answer: $x^{2} - 4 x y - y^{2} + 10 x + 2 y = c^{'}$ , where $c^{'} = - 2 c$

9.3 Exact equations

Solved Example 12

Solve

y d x + x d y = \frac{x d y - y d x}{x ^{2} + y ^{2}}

Solution:

The left side is $d (x y)$ and the right side is $d (tan^{- 1} \frac{y}{x})$ .
Integrate both sides.

Answer: $x y = tan^{- 1} \frac{y}{x} + c$

Solved Example 13

Solve

(2 x ln y) d x + (\frac{x ^{2}}{y} + 3 y^{2}) d y = 0

Solution:

Test: $\frac{\partial}{\partial y} (2 x ln y) = \frac{2 x}{y} = \frac{\partial}{\partial x} (\frac{x ^{2}}{y} + 3 y^{2})$ : exact (Figure 8).
Write it as $ln y (2 x d x) + x^{2} (\frac{d y}{y}) + 3 y^{2} d y = 0$ , i.e. $ln y d (x^{2}) + x^{2} d (ln y) + d (y^{3}) = 0$ .
So $d (x^{2} ln y) + d (y^{3}) = 0$ ; integrate each term.

Answer: $x^{2} ln y + y^{3} = c$

9.4 Linear, reducible to linear and Bernoulli

Solved Example 14

Solve

\frac{d y}{d x} + \frac{3 x ^{2}}{1 + x ^{3}} y = \frac{sin ^{2} x}{1 + x ^{3}}

Solution:

Linear with $P = \frac{3 x ^{2}}{1 + x ^{3}}$ : $IF = e^{\int P d x} = e^{l n (1 + x^{3})} = 1 + x^{3}$ .
$y (1 + x^{3}) = \int \frac{sin ^{2} x}{1 + x ^{3}} (1 + x^{3}) d x + c = \int \frac{1 - cos 2 x}{2} d x + c$ .

Answer: $y (1 + x^{3}) = \frac{x}{2} - \frac{sin 2 x}{4} + c$

Solved Example 15

Solve

x ln x \frac{d y}{d x} + y = 2 ln x

Solution:

Divide by $x ln x$ : $\frac{d y}{d x} + \frac{1}{x ln x} y = \frac{2}{x}$ , so $P = \frac{1}{x ln x}$ , $Q = \frac{2}{x}$ .
$IF = e^{\int P d x} = e^{l n (l n x)} = ln x$ .
$y ln x = \int \frac{2}{x} ln x d x + c$ .

Answer: $y ln x = (ln x)^{2} + c$

Solved Example 16

Solve

t (1 + t^{2}) d x = (x + x t^{2} - t^{2}) d t

, given that

x = - \frac{π}{4}

t = 1

Solution:

$t (1 + t^{2}) d x = [x (1 + t^{2}) - t^{2}] d t$ , so $\frac{d x}{d t} - \frac{x}{t} = - \frac{t}{1 + t ^{2}}$ : linear in $x$ .
$P = - \frac{1}{t}$ , $Q = - \frac{t}{1 + t ^{2}}$ , $IF = e^{- \int d t / t} = e^{- l n t} = \frac{1}{t}$ .
$x \cdot \frac{1}{t} = \int \frac{1}{t} (- \frac{t}{1 + t ^{2}}) d t + c = - tan^{- 1} t + c$ .
At $t = 1$ , $x = - \frac{π}{4}$ : $- \frac{π}{4} = - \frac{π}{4} + c$ , so $c = 0$ .

Answer: $x = - t tan^{- 1} t$

Solved Example 17

Solve

y sin x \frac{d y}{d x} = cos x (sin x - y^{2})

Solution:

Put $y^{2} = z$ : $2 y \frac{d y}{d x} = \frac{d z}{d x}$ . The equation becomes $\frac{sin x}{2} \frac{d z}{d x} + z cos x = sin x cos x$ .
$\frac{d z}{d x} + 2 cot x \cdot z = 2 cos x$ : linear in $z$ , $IF = e^{\int 2 c o t x d x} = e^{2 l n s i n x} = sin^{2} x$ .
$z sin^{2} x = \int 2 cos x sin^{2} x d x + c = \frac{2}{3} sin^{3} x + c$ .

Answer: $y^{2} sin^{2} x = \frac{2}{3} sin^{3} x + c$

Solved Example 18

Solve the Bernoulli equation

\frac{d y}{d x} - \frac{y}{x} = \frac{y ^{2}}{x ^{2}}

Solution:

Divide by $y^{2}$ : $\frac{1}{y ^{2}} \frac{d y}{d x} - \frac{1}{x y} = \frac{1}{x ^{2}}$ ... (1).
Put $\frac{1}{y} = t$ : $- \frac{1}{y ^{2}} \frac{d y}{d x} = \frac{d t}{d x}$ , so (1) becomes $- \frac{d t}{d x} - \frac{t}{x} = \frac{1}{x ^{2}}$ , i.e. $\frac{d t}{d x} + \frac{t}{x} = - \frac{1}{x ^{2}}$ .
Linear in $t$ : $IF = e^{\int d x / x} = x$ , and $t x = \int - \frac{1}{x} d x + c = - ln x + c$ .

Answer: $\frac{x}{y} = - ln x + c$

9.5 Clairaut's equation

Solved Example 19

Solve

y = m x + m - m^{3}

, where

m = \frac{d y}{d x}

Solution:

Clairaut form. Differentiate: $m = m + x \frac{d m}{d x} + \frac{d m}{d x} - 3 m^{2} \frac{d m}{d x}$ , so $\frac{d m}{d x} (x + 1 - 3 m^{2}) = 0$ .
$\frac{d m}{d x} = 0$ gives $m = c$ and the general solution $y = c x + c - c^{3}$ .
$x + 1 - 3 m^{2} = 0$ gives $m^{2} = \frac{x + 1}{3}$ . Then
$y = m (x + 1 - m^{2}) = \pm (\frac{x + 1}{3})^{1/2} \cdot \frac{2}{3} (x + 1) = \pm 2 (\frac{x + 1}{3})^{3/2}$
so $y^{2} = \frac{4}{27} (x + 1)^{3}$ .

Answer: general $y = c x + c - c^{3}$ ; singular $27 y^{2} = 4 (x + 1)^{3}$ (Figure 10)

9.6 Orthogonal trajectories

Solved Example 20

Find the orthogonal trajectory of the family of straight lines passing through the origin.

Solution:

Family: $y = m x$ ... (i); differentiating, $\frac{d y}{d x} = m$ ... (ii).
Eliminate $m$ : $y = x \frac{d y}{d x}$ .
Replace $\frac{d y}{d x}$ by $- \frac{d x}{d y}$ : $y = - x \frac{d x}{d y}$ , i.e. $x d x + y d y = 0$ .
Integrate: $\frac{x ^{2}}{2} + \frac{y ^{2}}{2} = c$ .

Answer: $x^{2} + y^{2} = 2 c$ , circles centred at the origin

Solved Example 21

Find the orthogonal trajectory of

y^{2} = 4 a x

(

a

being the parameter).

Solution:

$y^{2} = 4 a x$ ... (i); differentiating, $2 y \frac{d y}{d x} = 4 a$ ... (ii).
Eliminate $a$ : $y^{2} = 2 y \frac{d y}{d x} x$ , i.e. $y = 2 x \frac{d y}{d x}$ .
Replace $\frac{d y}{d x}$ by $- \frac{d x}{d y}$ : $y = - 2 x \frac{d x}{d y}$ , i.e. $2 x d x + y d y = 0$ .
Integrate: $x^{2} + \frac{y ^{2}}{2} = c$ .

Answer: $2 x^{2} + y^{2} = 2 c$ , a family of ellipses (Figure 11)

9.7 Geometrical applications

Solved Example 22

Find the nature of the curve for which the length of the normal at a point

P

is equal to the radius vector of

P

Solution:

Let $P (x, y)$ lie on $y = f (x)$ with slope $m = \frac{d y}{d x}$ . The normal at $P$ is $Y - y = - \frac{1}{m} (X - x)$ , meeting the $x$ -axis at $G (x + m y, 0)$ (Figure 12).
$O P^{2} = P G^{2}$ : $x^{2} + y^{2} = m^{2} y^{2} + y^{2}$ , so $m = \pm \frac{x}{y}$ .
Taking $+$ : $y d y = x d x$ , so $\frac{y ^{2}}{2} = \frac{x ^{2}}{2} + λ$ , i.e. $x^{2} - y^{2} = c$ .
Taking $-$ : $y d y = - x d x$ , so $x^{2} + y^{2} = c^{'}$ .

Answer: a rectangular hyperbola $x^{2} - y^{2} = c$ or a circle $x^{2} + y^{2} = c^{'}$

Solved Example 23

Find the curves for which the portion of the tangent included between the coordinate axes is bisected at the point of contact.

Solution:

Tangent at $P (x, y)$ : $Y - y = m (X - x)$ . It meets the axes at $A (\frac{m x - y}{m}, 0)$ and $B (0, y - m x)$ .
$P$ is the midpoint of $A B$ : $\frac{m x - y}{m} = 2 x$ , so $m x - y = 2 m x$ , i.e. $x \frac{d y}{d x} = - y$ .
Separate: $\frac{d x}{x} + \frac{d y}{y} = 0$ , so $ln x + ln y = ln c$ .

Answer: $x y = c$ , rectangular hyperbolas (Figure 14)

Figure 14: For

x y = c

the tangent at

P (x, y)

meets the axes at

A (2 x, 0)

and

B (0, 2 y)

, so

P

is always the midpoint. Drawn for

c = 2

P (1, 2)

Solved Example 24

Show that

(4 x + 3 y + 1) d x + (3 x + 2 y + 1) d y = 0

represents a hyperbola having the lines

x + y = 0

and

2 x + y + 1 = 0

as asymptotes.

Solution:

Regroup: $4 x d x + 3 (y d x + x d y) + d x + 2 y d y + d y = 0$ . Integrating each term: $2 x^{2} + 3 x y + y^{2} + x + y + c = 0$ .
Here $a = 2$ , $h = \frac{3}{2}$ , $b = 1$ , so $h^{2} = \frac{9}{4} > ab = 2$ : a hyperbola (for $Δ \neq = 0$ ).
Asymptotes: $2 x^{2} + 3 x y + y^{2} + x + y + λ = 0$ must be a pair of lines, so $Δ = 0$ with $g = f = \frac{1}{2}$ : $2 λ + 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} - 2 \cdot \frac{1}{4} - 1 \cdot \frac{1}{4} - \frac{9}{4} λ = 0$ , so $λ = 0$ .
$2 x^{2} + 3 x y + y^{2} + x + y = (x + y) (2 x + y) + (x + y) = (x + y) (2 x + y + 1)$ .

Answer: asymptotes $x + y = 0$ and $2 x + y + 1 = 0$ (the curve is a hyperbola for $c \neq = 0$ )

Solved Example 25

The perpendicular from the origin to the tangent at any point of a curve is equal to the abscissa of the point of contact. Find the curve that satisfies this and passes through

(1, 1)

Solution:

Tangent at $P (x, y)$ : $m X - Y + y - m x = 0$ . Its distance from $O$ : $\frac{∣ y - m x ∣}{1 + m ^{2}} = x$ .
Square: $y^{2} + m^{2} x^{2} - 2 m x y = x^{2} (1 + m^{2})$ , so $\frac{d y}{d x} = \frac{y ^{2} - x ^{2}}{2 x y}$ , a homogeneous equation.
$y = v x$ : $v + x \frac{d v}{d x} = \frac{v ^{2} - 1}{2 v}$ , so $x \frac{d v}{d x} = - \frac{v ^{2} + 1}{2 v}$ .
$\int \frac{2 v}{v ^{2} + 1} d v = - \int \frac{d x}{x}$ : $ln (v^{2} + 1) = - ln x + ln c$ , so $x (\frac{y ^{2}}{x ^{2}} + 1) = c$ .
Through $(1, 1)$ : $c = 2$ .

Answer: $x^{2} + y^{2} - 2 x = 0$ , the circle of Solved Example 8

9.8 Exam-style problems

Solved Example 26

(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2}

and

y (0) = 0

, then

y (1)

equals
(A)

1/3

(B)

2/3

(C)

4/3

(D)

1

Solution:

The left side is $\frac{d}{d x} [y (1 + x^{2})]$ (the IF is already in place). So $y (1 + x^{2}) = \frac{4 x ^{3}}{3} + c$ , and $y (0) = 0$ gives $c = 0$ . Then $y = \frac{4 x ^{3}}{3 ( 1 + x ^{2} )}$ and $y (1) = \frac{4}{3} \cdot \frac{1}{2}$ .

Answer: (B) $\frac{2}{3}$

Solved Example 27

\frac{d y}{d x} = \frac{1 + y ^{2}}{1 + x ^{2}}

and

y (0) = 1

, then

y (\frac{1}{2})

equals
(A)

2

(B)

3

(C)

1/3

(D)

4

Solution:

Separate: $tan^{- 1} y = tan^{- 1} x + c$ ; $y (0) = 1$ gives $c = \frac{π}{4}$ . So $y = tan (tan^{- 1} x + \frac{π}{4}) = \frac{1 + x}{1 - x}$ (for $x < 1$ ), and $y (\frac{1}{2}) = \frac{3/2}{1/2} = 3$ .

Answer: (B) $3$

Solved Example 28

Solve

(1 + e^{x / y}) d x + e^{x / y} (1 - \frac{x}{y}) d y = 0

Solution:

$\frac{d x}{d y} = - \frac{e ^{x / y} ( 1 - x / y )}{1 + e ^{x / y}}$ , a function of $\frac{x}{y}$ : put $x = v y$ , $\frac{d x}{d y} = v + y \frac{d v}{d y}$ .
$y \frac{d v}{d y} = - \frac{e ^{v} ( 1 - v )}{1 + e ^{v}} - v = - \frac{v + e ^{v}}{1 + e ^{v}}$ .
$\int \frac{1 + e ^{v}}{v + e ^{v}} d v = - \int \frac{d y}{y}$ : $ln (v + e^{v}) = - ln y + ln c$ , so $y (v + e^{v}) = c$ .

Answer: $x + y e^{x / y} = c$

Solved Example 29

Solve

y^{2} d x + (x - \frac{1}{y}) d y = 0

Solution:

Not linear in $y$ , but $\frac{d x}{d y} + \frac{x}{y ^{2}} = \frac{1}{y ^{3}}$ is linear in $x$ .
$IF = e^{\int y^{- 2} d y} = e^{- 1/ y}$ , so $x e^{- 1/ y} = \int \frac{e ^{- 1/ y}}{y ^{3}} d y$ .
Put $t = \frac{1}{y}$ , $d t = - \frac{d y}{y ^{2}}$ : the integral is $- \int t e^{- t} d t = (t + 1) e^{- t} + c$ .
So $x e^{- 1/ y} = (\frac{1}{y} + 1) e^{- 1/ y} + c$ .

Answer: $x = 1 + \frac{1}{y} + c e^{1/ y}$

Solved Example 30

Find the orthogonal trajectories of the family of parabolas

y = a x^{2}

Solution:

$\frac{d y}{d x} = 2 a x$ and $a = \frac{y}{x ^{2}}$ , so $\frac{d y}{d x} = \frac{2 y}{x}$ .
Replace $\frac{d y}{d x}$ by $- \frac{d x}{d y}$ : $- \frac{d x}{d y} = \frac{2 y}{x}$ , i.e. $x d x = - 2 y d y$ .
Integrate: $\frac{x ^{2}}{2} = - y^{2} + c$ .

Answer: $x^{2} + 2 y^{2} = c$ , a family of ellipses

Practice Questions

Separable: $x^{2} y \frac{d y}{d x} = (x + 1) (y + 1)$ Answer: $y - ln (y + 1) = ln x - \frac{1}{x} + c$
Separable: $\frac{x d x + y d y}{x ^{2} + y ^{2}} = \frac{y d x - x d y}{x ^{2}}$ Answer: $x^{2} + y^{2} + \frac{y}{x} = c$
Separable: $\frac{d y}{d x} = e^{x + y} + x^{2} e^{y}$ Answer: $- e^{- y} = e^{x} + \frac{x ^{3}}{3} + c$
Separable: $x y \frac{d y}{d x} = 1 + x + y + x y$ Answer: $y = x + ln ∣ x (1 + y) ∣ + c$
Reducible: $\frac{d y}{d x} = 1 + e^{x - y}$ Answer: $e^{y - x} = x + c$
Reducible: $\frac{d y}{d x} = sin (x + y) + cos (x + y)$ Answer: $ln tan \frac{x + y}{2} + 1 = x + c$
Reducible: $\frac{d y}{d x} = x tan (y - x) + 1$ Answer: $sin (y - x) = e^{x^{2} /2 + c}$
Homogeneous: $(x \frac{d y}{d x} - y) tan^{- 1} \frac{y}{x} = x$ , given $y = 0$ at $x = 1$ Answer: $x^{2} + y^{2} = e^{(y / x) t a n^{- 1} (y / x)}$
Homogeneous: $x \frac{d y}{d x} = y - x tan \frac{y}{x}$ Answer: $x sin \frac{y}{x} = C$
Shift of origin: $\frac{d y}{d x} = \frac{x + 2 y - 3}{2 x + y + 3}$ Answer: $x + y = c (x - y + 6)^{3}$
Case 1: $\frac{d y}{d x} = \frac{x + y + 1}{2 x + 2 y + 3}$ Answer: $3 (2 y - x) + ln (3 x + 3 y + 4) = C$
Case 2: $\frac{d y}{d x} = \frac{3 x + 2 y - 5}{3 y - 2 x + 5}$ Answer: $3 x^{2} + 4 x y - 3 y^{2} - 10 x - 10 y = C$
Exact: $x d y + y d x + x y e^{y} d y = 0$ Answer: $ln (x y) + e^{y} = c$
Exact: $y e^{- x / y} d x - (x e^{- x / y} + y^{3}) d y = 0$ Answer: $2 e^{- x / y} + y^{2} = c$
Linear: $x (x^{2} + 1) \frac{d y}{d x} = y (1 - x^{2}) + x^{2} ln x$ Answer: $(\frac{x ^{2} + 1}{x}) y = x ln x - x + c$
Linear in x: $(x + 2 y^{3}) \frac{d y}{d x} = y$ Answer: $x = y (c + y^{2})$
Bernoulli: $x \frac{d y}{d x} + y = y^{2} ln x$ Answer: $y (1 + c x + ln x) = 1$
Change of variable: $x y^{2} \frac{d y}{d x} - 2 y^{3} = 2 x^{3}$ , given $y = 1$ at $x = 1$ Answer: $y^{3} + 2 x^{3} = 3 x^{6}$
Clairaut: $y = m x + \frac{2}{m}$ , where $m = \frac{d y}{d x}$ Answer: general $y = c x + \frac{2}{c}$ ; singular $y^{2} = 8 x$
Clairaut: $sin p x cos y = cos p x sin y + p$ , where $p = \frac{d y}{d x}$ Answer: general $y = c x - sin^{- 1} c$ ; singular $y = x^{2} - 1 - sin^{- 1} \frac{x ^{2} - 1}{x}$
Orthogonal: family of circles concentric at $(a, 0)$ Answer: $y = c (x - a)$
Orthogonal: family of circles touching the $x$ -axis at the originAnswer: $x^{2} + y^{2} = c x$
Orthogonal: rectangular hyperbolas $x y = c^{2}$ Answer: $x^{2} - y^{2} = k$

Common Mistakes to Avoid

Watch out

Dividing by a factor that can be zero and losing a solution: $\frac{d y}{d x} = x y$ also has $y = 0$ (Figure 2).
Writing $\frac{d y}{d x} = v + \frac{d v}{d x}$ after $y = v x$ . The product rule gives $v + x \frac{d v}{d x}$ .
Finding the IF before making the coefficient of $\frac{d y}{d x}$ equal to 1. In $x ln x \frac{d y}{d x} + y = 2 ln x$ divide by $x ln x$ first.
Adding a constant inside the IF, or forgetting the $+ C$ after $\int Q \cdot IF d x$ .
Leaving the answer in $X$ , $Y$ , $v$ , $t$ or $u$ . Always substitute back ( $X = x - h$ , $Y = y - k$ , $v = \frac{y}{x}$ ).
Testing exactness with $\frac{\partial M}{\partial x}$ and $\frac{\partial N}{\partial y}$ . Each coefficient is differentiated with respect to the other variable.
In orthogonal trajectories, replacing $\frac{d y}{d x}$ by $- \frac{d x}{d y}$ before the constant is eliminated.
In Bernoulli's equation, using $z = y^{n - 1}$ instead of $y^{1 - n}$ , or dropping the factor $(1 - n)$ .

Frequently Asked Questions

What are the methods of solving first order first degree differential equations?

The standard methods are variables separable, homogeneous equations solved with $y = v x$ , exact equations, and linear equations solved with an integrating factor. Substitutions such as $t = a x + b y + c$ , a shift of origin, polar coordinates and Bernoulli's $z = y^{(1 - n)}$ reduce other equations to one of these four types.

How do you know if a differential equation is homogeneous?

Write it as $\frac{d y}{d x} = \frac{f ( x , y )}{g ( x , y )}$ . It is homogeneous if f and g are homogeneous functions of the same degree, which means replacing x by kx and y by ky multiplies both by the same power of k. Then dy/dx depends only on y/x and the substitution y = vx separates the variables.

What is an integrating factor in a linear differential equation?

For $\frac{d y}{d x} + P y = Q$ , the integrating factor is e raised to the integral of P dx. Multiplying by it turns the left side into the derivative of y times the factor, so y times IF equals the integral of Q times IF plus C. No constant is added while finding the factor itself.

How do you check whether a differential equation is exact?

Write the equation as M dx + N dy = 0. It is exact when the partial derivative of M with respect to y equals the partial derivative of N with respect to x. Then M dx + N dy is the total differential of some F(x, y), and the solution is F(x, y) = c.

What is Bernoulli's differential equation and how is it solved?

Bernoulli's equation is $\frac{d y}{d x} + P y = Q y^{n}$ with n not equal to 0 or 1, where P and Q depend on x only. Divide by $y^{n}$ and substitute $z = y^{(1 - n)}$ . The equation becomes $\frac{d z}{d x} + (1 - n) P z = (1 - n) Q$ , which is linear in z and is solved with the integrating factor.

When should x be taken as the dependent variable in a differential equation?

When the equation is not linear in y but x appears only to the first power, rewrite it for dx/dy. For example, $(x + 2 y^{3}) \frac{d y}{d x} = y$ becomes $\frac{d x}{d y} - \frac{x}{y} = 2 y^{2}$ , a linear equation in x with integrating factor e raised to the integral of P1 dy, here 1/y.

What is the singular solution of Clairaut's equation?

Clairaut's equation $y = p x + f (p)$ , with $p = \frac{d y}{d x}$ , has the general solution $y = c x + f (c)$ , a family of straight lines. Eliminating p between the equation and $x + f^{'} (p) = 0$ gives the singular solution, which contains no constant and is the envelope that touches every one of those lines.

Which differential equation methods are asked in JEE Main?

The JEE Main syllabus names solution by separation of variables, homogeneous equations and linear equations of the type $\frac{d y}{d x} + P (x) y = Q (x)$ , so these three appear most often, usually with an initial condition. Exact equations, Bernoulli's equation, substitutions and orthogonal trajectories are common in JEE Advanced level problems.

Previous year questions on Methods of Solving First Order, First Degree Differential Equation

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1. First Order, First Degree Differential Equations

2. Variables Separable

2.1 The method

2.2 Polar coordinate substitution

2.3 Equations reducible to variables separable

3. Homogeneous Differential Equations

3.1 Homogeneous functions

3.2 Solving by y=vx

3.3 Equations reducible to homogeneous form

4. Exact Differential Equations

5. Linear Differential Equations

5.1 What makes an equation linear

5.2 First order linear equation and the integrating factor

5.3 Linear in x

5.4 Reducible to linear by a change of variable

5.5 Bernoulli's equation

6. Clairaut's Equation

7. Orthogonal Trajectories

8. Geometrical Applications

9. Solved Examples

9.1 Variables separable and reducible forms

9.2 Homogeneous and reducible to homogeneous

9.3 Exact equations

9.4 Linear, reducible to linear and Bernoulli

9.5 Clairaut's equation

9.6 Orthogonal trajectories

9.7 Geometrical applications

9.8 Exam-style problems

Common Mistakes to Avoid

Frequently Asked Questions

What are the methods of solving first order first degree differential equations?

How do you know if a differential equation is homogeneous?

What is an integrating factor in a linear differential equation?

How do you check whether a differential equation is exact?

What is Bernoulli's differential equation and how is it solved?

When should x be taken as the dependent variable in a differential equation?

What is the singular solution of Clairaut's equation?

Which differential equation methods are asked in JEE Main?

Previous year questions on Methods of Solving First Order, First Degree Differential Equation

Ready to master Differential Equations?

3.2 Solving by $y = v x$

5.3 Linear in $x$