Methods of Solving First Order, First Degree Differential Equation

MathsDifferential EquationsFor JEE aspirants

SOLUTION OF DIFFERENTIAL EQUATION

The general solution of a differential equation is the relation in the variables x, y obtained by integrating (removing derivatives ) where the relation contains as many arbitrary constants as the order of the equation. The general solution of a differential equation of the first order contains one arbitrary constant while that of the second order contains two arbitrary constants. In the general solution, if particular values of the arbitrary constant are put , we get a particular solution which will give one member of the family of curves.

To solve differential equation of the first order and the first degree:

Simple standard form of differential equation of the first order and first degree are as follows:

(i) Variable Separable

Form f(x) dx + $ϕ$ y) dy = 0

Method: Integrate it i.e., find $\int f (x) d x + \int ϕ (y) dy = c$

Illustration 1: Solve $\frac{d y}{d x} = e^{x} y$ .

Solution: Given $\frac{d y}{d x} = e^{x} y$ $\Rightarrow$ $\frac{d y}{d x} = e^{x} y$

$\Rightarrow$ $\frac{d y}{y} - e^{x} d x = 0$

Integrating, we get ln y – e^x =c

(ii) Reducible into Variable Separable

Method: Sometimes differential equation of the first order cannot be solved directly by variable separation but by some substitution we can reduce it to a differential equation with separable variables.

A differential equation of the $\frac{d y}{d x} = f (a x + b y + c)$ is solved by writing ax + by + c = t

Illustration 2: Solve (x – y)² $\frac{d y}{d x} = 1$ .

Solution: Put z = x –y $\Rightarrow$ $\frac{d z}{d x} = 1 - \frac{d y}{d x}$ $\Rightarrow$ $\frac{d y}{d x} = 1 - \frac{d z}{d x}$

Now z^{2 $(1 - \frac{d z}{d x}) = 1$} $\Rightarrow$ $z^{2} \frac{d z}{d x} = z^{2} - 1$

$\Rightarrow$ dx = $\frac{z ^{2}}{z ^{2} - 1} d z$ , which is in the form of variable separable

Now integrating, we get x = z + $\frac{1}{2} ln \frac{z - 1}{z + 1} + c$

$\Rightarrow$ Solution is x =(x – y) + $\frac{1}{2} ln \frac{x - y - 1}{x + y + 1} + 1$

(iii) Homogeneous Equation

When $\frac{d y}{d x}$ is equal to a fraction whose numerator and the denominator both are homogeneous functions of x and y of the same degree then the differential equation is said to be homogeneous equation.

i.e. when $\frac{d y}{d x} = f (x, y)$ , where f(km, ky) = f(x, y) then this differential equation is said to be homogeneous differential equation .

Method: Put y = vx

Illustration 3: Solve $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{2 x y}$ .

Solution: $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{2 x y}$ (homogeneous ) . Put y = vx

$∴ \frac{x ^{2} + v ^{2} x ^{2}}{2 x . v x} = v + x \frac{d v}{d x}$

$\Rightarrow$ $\frac{d x}{x} = \frac{2 v}{1 - v ^{2}} d v$ , Integrate

C + lnx = - ln(1 –v²)

$\Rightarrow$ lnkx + ln(1 –v²) =0

$\Rightarrow$ kx(1- v²) = 1 $\Rightarrow$ k(x² – y²) = x .

(iv) Non-homogeneous Differential Equation

Form $\frac{d y}{d x} = \frac{a _{1} x + b _{1} y + c _{1}}{a _{2} x + b _{2} y + c _{2}}$

Method: If $\frac{a _{1}}{a _{2}} \neq = \frac{b _{1}}{b _{2}}$ , put x = X + h , y = Y + k such that a₁h +b₁k + c₁ = 0, a₂h + b₂k +c₂ = 0 . In this way the equation becomes homogeneous in X, Y. then use the method for homogeneous equation.

If $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}}$ . Put a₁x+b₁y=v or a₂x + b₂y = v. The equation in the form of variable separable in x, v.

Illustration 4: $\frac{d y}{d x} = \frac{x - y + 3}{2 x - 2 y + 5}$ .

Solution: Here $\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{1}{2}$

Hence we put x- y = v

$1 - \frac{d y}{d x} = \frac{d v}{d x}$ $\Rightarrow$ $\frac{d y}{d x} = 1 - \frac{d y}{d x}$

or, 1 – $\frac{d v}{d x} = \frac{v + 3}{2 v + 5}$ $\Rightarrow$ $\frac{2 v + 5}{v + 2} d v$ = dx or, $(2 + \frac{1}{v + 2}) d v = d x$ Integrate

2v + ln (v +2) = x + C, Put the value of v

$∴$ x – 2y+ ln (x – y +2) = C(v) Linear Equation Form $\frac{d y}{d x} + P (x) y = Q (x)$ , where P(x)andQ(x) arefunctionsof xMethod: Multiplying the equation by $e^{\int P (x) d x}$ , called integrating factor. Then the equation becomes $\frac{d}{d x} {y . e^{\int P (x) d x}} = Q (x) e^{\int P (x) d x}$ Integrating $y . e^{\int P (x) d x} = \int {Q (x) e^{\int P (x) d x}} d x$ (vi) Reducible into Linear Equation FormR(y) $\frac{d y}{d x}$ + P(x) S(y) = Q(x) , suchthat $\frac{d [ S ( y ) ]}{d y} = R (y)$ Method: PutS(y) =zthen $\frac{d z}{d x} = \frac{d [ S ( y ) ]}{d x} = R (y) \frac{d y}{d x}$ $∴$ The;equation;becomes; $\frac{d z}{d x} + p (x) z = Q (x)$ ,;which;is in the;linear;form

Illustration 5:; $\frac{d y}{d x} + \frac{y}{( 1 + x ^{2} )} = \frac{e ^{t a n^{- 1} x}}{1 + x ^{2}}$ .Solution:;;;;;;;; $\frac{d y}{d x} + \frac{y}{( 1 + x ^{2} )} = \frac{e ^{t a n^{- 1} x}}{1 + x ^{2}}$ ;;;;;;;;;;;;;;;;;;;;;;;Multiplying;both sides by;I.F. and;integrating;;;;;;;;;;;;;;;;;;;;;;; $y e^{t a n^{- 1} x} = \int \frac{e tan ^{- 1} x}{( 1 + x ^{2} )} e^{t a n^{- 1} x} d x + c$ ;;;;;;;;;;;;;;;;;;;;;;;;Put; $e^{t a n^{- 1} x} = t, \frac{e ^{t a n^{- 1} x}}{1 + x ^{2}} d x = d t$ ;;;;;;;;;;;;;;;;;;;;;;; $\Rightarrow$ $y e^{t a n^{- 1} x} = \int t d t + c = \frac{t ^{2}}{2} + c$ ;;;;;;;;;;;;;;;;;;;;;;;; $\Rightarrow$ 2 y $e^{t a n^{- 1} x} = e^{2 t a n^{- 1} x} + 2 c$

;(vii) ;;;Exact Differential Equations Mdx + Ndy = 0, where;M and N are;functions of;x and y. If; $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ , then the;equation;is exact;and;its;solution is;given;by;ò Mdx + ò N dy;= cTo find the solution of an exact differential equation Mdx + N dy = 0, integrate $\int M d x$ as if y were constant. Also integrate the terms of N that do not contain x w.r.t y. Equate the sum of these integrals to a constant.

Illustration 6:;(x² –ay)dx + (y² –ax)dy = 0.Solution:;;;;;;;;Here M = x² –ay;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;N = y² –ax;;;;;;;;;;;;;;;;;;;;;;; $\frac{\partial M}{\partial y} = - a$ ;;;;;;;;;;;;;;;;;;;;;;; $\frac{\partial N}{\partial x} = - a$ $\Rightarrow$ $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ ;;;;;;;;;;;;;;;;;;;;;;; $∴$ equation is exact ;;;;;;;;;;;solution is $\int (x^{2} - a y) d x + \int y^{2} d y$ = c;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;; $\frac{x ^{3}}{3}$ – ayx + $\frac{y ^{3}}{3}$ = c;;;;;;;;;;;;;;;;;;;;;;;or x³ –3axy + y³ = 3c.;Integrating Factor: A factor, which when multiplied to a non exact, differential equation makes it exact, is known as integrating factor e.g. the non exact equation y dx –x dy = 0 can be made exact on multiplying by the factor $\frac{1}{y ^{2}}$ . Hence $\frac{1}{y ^{2}}$ is the integrating factor for this equation.

Notes:In general such a factor exist but except in certain special cases, it is likely to be difficult to determine.The number of integrating factor for equation M dx + N dy = 0 is infinite.

Some Useful Results:;d(xy) = xdy + ydx $d (\frac{y}{x}) = \frac{x d y - y d x}{x ^{2}}$ $d (\frac{x}{y}) = \frac{y d x - x d y}{y ^{2}}$ $d [ln (\frac{y}{x})]$ =; $\frac{x d y - y d x}{x y}$ d tan^-1 $(\frac{y}{x})$ = $\frac{x d y - y d x}{x ^{2} + y ^{2}}$ $d tan^{- 1} (\frac{x}{y}) = \frac{y d x}{x ^{2} + y ^{2}} .$ d(sin^-1 xy)= $\frac{x d y + y d x}{1 - ( x y ) ^{2}}$ Illustration 7:Solvex dy – y dx= $x^{2} + y^{2} d x$ .Solution: $\frac{x d y - y d x}{x ^{2}} = 1 + (\frac{y}{x})^{2} \frac{d x}{x}$ $\Rightarrow$ $\frac{1}{1 + ( \frac{y}{x} ) ^{2}} (\frac{x d y - y d x}{x ^{2}}) = \frac{d x}{x}$ $\Rightarrow$ $\frac{1}{1 + ( \frac{y}{x} ) ^{2}} d (\frac{y}{x}) = \frac{d x}{x}$ Integrating $\Rightarrow$ ln $(\frac{y}{x} + 1 + \frac{y ^{2}}{x ^{2}}) = ln x + ln k$ $\Rightarrow$ ln $(y + x^{2} + y^{2})$ = 2 ln x + ln ky+ $x^{2} + y^{2}$ = kx². (viii) Linear Differential Equation with constant coefficient Differential equation of the form $a_{0} \frac{d ^{n} y}{d x ^{n}} + a_{1} \frac{d ^{n - 1} y}{d x ^{n - 1}} + .... + a_{n} y = 0$ .,a_I $\in$ R forI =0, 1 , 2, 3, . . . , niscalleda lineardifferential equationwithconstantcoefficients.In order to solve this differential equation, take the auxiliary equation as a₀Dⁿ + a₁D^n-1 +…+ a_n =0Find the roots of this equation and then solution of the given differential equation will be as given in the following table

Diagram being restored — will be back shortly

Illustration 8:Solve $\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 = 0$ .Solution: Itsauxiliaryequation is D² � + 2 = 0

\Rightarrow

D = 1,D = 2Hence its solution isy = C₁e^x + C₂e^2xSo far only linear differential equation with constant coefficients of forma₀

\frac{d ^{n} y}{d x ^{n}}

+ a₁

\frac{d ^{n - 1} y}{d x ^{n - 1}}

..+ ….+ a_n y = 0,a_I

\in

Rfori = 0, 1, 2, …., n were considered. Now we consider the following forma₀

\frac{d ^{n} y}{d x ^{n}}

+ a₁

\frac{d ^{n - 1} y}{d x ^{n - 1}}

.+ ….+ a_n y = Xwhere X is either constant or functions of x alone. Theorem:Ify = f₁(x) is the general solution of a₀

\frac{d ^{n} y}{d x ^{n}}

+ a₁

\frac{d ^{n - 1} y}{d x ^{n - 1}}

.+ ….+ a_n y = 0andy = f₂(x)is a solution ofa₀

\frac{d ^{n} y}{d x ^{n}}

+ a₁

\frac{d ^{n - 1} y}{d x ^{n - 1}}

.+ ….+ a_n y = XThen y = f₁(x) + f₂(x) is the general solution of a₀

\frac{d ^{n} y}{d x ^{n}}

+ a₁

\frac{d ^{n - 1} y}{d x ^{n - 1}}

+ ….+ a_ny = XExpression f₁(x)is known as complementary function and the expression f₂(x) is known as particular integral. These are denotes by C.F.and P.I. respectively.The n^th derivative of y will be denoted Dⁿy where D stands for

\frac{d}{d x}

and n denotes the order of derivative.If we take Differential Equation:

\frac{d ^{n} y}{d x ^{n}}

+ P₁

\frac{d ^{n - 1} y}{d x ^{n - 1}}

+ P₂

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

+ …. + P_ny = Xthen we can write this differential equation in a symbolic form asDⁿy + P₁D^{n– 1}y + P₂D^{n– 2}y + ….+ P_ny = X(Dⁿ + P₁D^{n– 1} + P₂D^{n– 2} + ….+ P_n)y = XThe operator Dⁿ + P₁D^{n– 1} + P₂D^{n– 2} + ….+ P_n is denoted by f(D) so that the equation takes the form f(D)y = Xy =

\frac{1}{f ( D )} X

Methods of finding P.I. :Notes: If both m₁ and m₂ are constants, the expressions (D – m₁)(D – m₂)y and (D– m₂)(D – m₁)y are equivalent i.e. the expression is independent of the order of operational factors.

\frac{1}{D - α} X = e^{α x} \int X e^{- α x} d x

We will explain the method with the help of followingIllustration 9:Solve $\frac{d ^{2} y}{d x} - 5 \frac{d y}{d x} + 6 y = e^{3 x}$ .Solution:The equation can be written as (D² – 5D + 6)y = e^3x(D – 3) (D – 2)y = e^3xC.F.= c₁ e^3x + c₂e^2xAndP.I. =

\frac{1}{( D - 3 )} . \frac{1}{( D - 2 )} e^{3 x}

\frac{1}{( D - 3 )} e^{2 x} \int e^{3 x} e^{- 2 x} d x

\frac{1}{( D - 3 )} . e^{2 x} . e^{x}

= e^3x

\int e^{3 x} . e^{- 3 x} d x

= x. e^3x

∴

y = c₁ e^3x + c₂e^2x + x e^3xP.I. can be found by resolving

\frac{1}{f ( D )} = \frac{1}{( D - 3 )} . \frac{1}{( D - 2 )}

Into partial functions

\frac{1}{f ( D )} = \frac{1}{( D - 3 )} . \frac{1}{( D - 2 )} = \frac{1}{( D - 3 )} - \frac{1}{( D - 2 )}

∴

P.I.=

(\frac{1}{( D - 3 )} - \frac{1}{( D - 2 )}) e^{3 x}

\frac{1}{( D - 3 )} e^{3 x} - \frac{1}{( D - 2 )} e^{3 x}

e^{3 x} \int e^{3 x} . e^{- 3 x} d x - e^{2 x} \int e^{3 x} . e^{- 2 x} d x

= x e^3x 𠄾^3x .Second term can be neglected as it is included as it inclined in the first term of a C.F.Short Method of Finding P.I. : In certain cases, the P.I.can be obtained by methods shorter than the general method.(i). To find P.I.whenX = e^ax in f(D) y = X, wherea is constanty =

\frac{1}{f ( D )}

\frac{1}{f ( D )} e^{a x} = \frac{1}{f ( a )} e^{a x}

if f(a)

\neq =

\frac{1}{f ( D )} e^{a x} = \frac{x ^{r} e ^{a x}}{f ^{r} ( a )}

iff(a) = 0 , where f(D) = (D – a)^r

ϕ

(D)Illustration 10: Solve (D³ �² + 7D – 3)y = e^3x. Solution:(D 𠄱)² (D – 3) y = e^3xC.F.= ae^x + bx e^x + ce^3xAndP.I.=

\frac{1}{D ^{3} - 5 D ^{2} + 7 D - 3}

e^3x=

\frac{x}{3 D ^{2} - 10 D + 7} e^{3 x} = \frac{1}{4} x . e^{3 x}

∴

y = ae^x + bx e^x + ce^3x+

\frac{1}{4} x e^{3 x}

(ii). To find P.I.when X = cosax or sinaxf(D) y = Xy =

\frac{1}{f ( D )}

sinaxIff( – a²)

\neq =

0then

\frac{1}{f ( D ^{2} )} sin a x

\frac{1}{f ( - a ^{2} )} sin a x

If f(– a²) = 0 then (D² + a²) is atleast one factor of f(D²)Let f (D²) = (D² + a²)^r

ϕ

(D²)Where

ϕ

(– a²)

\neq =

∴

\frac{1}{f ( D ^{2} )} sin a x

\frac{1}{( D ^{2} + a ^{2} ) ^{r}} \frac{1}{φ ( D ^{2} )} sin a x

\frac{1}{φ ( - a ^{2} )} \frac{1}{( D ^{2} + a ^{2} ) ^{r}} sin a x

when r = 1

\frac{1}{D ^{2} + a ^{2}}

sin ax= –

\frac{x}{2 a} cos a x .

.Similarly Iff(– a)²

\neq =

0then

\frac{1}{f ( D ^{2} )}

..cos ax =

\frac{1}{f ( - a ^{2} )}

cosaxand

\frac{1}{D ^{2} + a ^{2}} cos a x = \frac{x}{2 a} sin a x

Illustration 11:Solve(D² � + 6)y = sin3x. Solution:(D 𠄲) (D – 3)y = sin3xC.F.= ae^2x + be^3xP.I.=

\frac{1}{D ^{2} - 5 D + 6}

sin3x =

\frac{1}{( - 9 ) - 5 D + 6} sin 3 x

\frac{1}{- 5 D - 3} sin 3 x

= – (5D – 3).

\frac{1}{25 D ^{2} - 9}

sin3x=

\frac{1}{234} (15 cos 3 x - 3 sin 3 x)

∴

ae^2x + be^3x +

\frac{1}{234} (15 cos 3 x - 3 sin 3 x)

(iii). To find the P.I.when X = x^mwhere m

\in

Nf(D) y = x^my =

\frac{1}{f ( D )} x^{m}

we will explain the method by taking an exampleIllustration 12: Find P.I. of (D³ + 3D² + 2D)y = x². Solution:P.I.=

\frac{1}{D ^{2} + 3 D ^{2} + 2 D} x^{2}

\frac{1}{2 D ( 1 + \frac{3 D + D ^{2}}{2} )} x^{2}

\frac{1}{2 D} (1 + \frac{3 D + D ^{2}}{2})^{- 1} x^{2}

\frac{1}{2 D} {1 - \frac{3 D + D ^{2}}{2} + \frac{9}{4} D^{2} + ....} x^{2}

\frac{1}{2 D} {1 - \frac{3 D}{2} + \frac{7}{4} D^{2} + ....} x^{2}

\frac{1}{2 D} {x^{2} - 3 x + \frac{7}{2}}

\frac{1}{2 D} {x^{2} - 3 x + \frac{7}{2}} = \frac{1}{2} {\frac{x ^{3}}{3} - \frac{3 x ^{2}}{2} + \frac{7}{2} x}

\frac{1}{12} {2 x^{3} - 9 x^{2} + 21 x}

(iv). To find the value of

\frac{1}{f ( D )}

e^axV where 𠆊’ is a constant and V is a function of x

\frac{1}{f ( D )} e^{a x} V = e^{a x} \frac{1}{f ( D + a )} V

Illustration 13: Solve(D² + 2)y = x² e^3x. Solution:C.F.= a cos

2

x + b sin

2

xP.I.=

\frac{1}{D ^{2} + 2}

x² e^3x = e^3x

\frac{1}{( D + 3 ) ^{2} + 2} x^{2}

e^{3 x}

\frac{1}{D ^{2} + 6 D + 11} x^{2}

\frac{1}{11} e^{3 x} (1 + \frac{6 D + D ^{2}}{11})^{- 1}

x²

= $\frac{1}{11} e^{3 x} (1 - \frac{6 D + D ^{2}}{11} + \frac{36}{121} D^{2} + ....)$ x²= $\frac{1}{11} e^{3 x} (1 - \frac{6 D}{11} + \frac{25}{121} D^{2} + ....)$ x²

= $\frac{1}{11} e^{3 x} (x^{2} - \frac{12}{11} x + \frac{50}{121})$

$∴$ a cos $2$ x + b sin $2$ x + $\frac{e ^{3 x}}{11} (x^{2} - \frac{12}{11} x + \frac{50}{121})$

(v). To find $\frac{1}{f ( D )} x V$ where V is a function of x

$\frac{1}{f ( D )} x V = [x - \frac{1}{f ( D )} f^{'} (D)] \frac{1}{f ( D )} V$

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