JEE Main 2025 Jan 24 Shift 1, Mathematics Q21: Methods of Solving First Order, First Degree Differential Equation

Differentiate both sides with respect to $x$ (Leibniz on the right gives $10(x+2)f(x)$):

$4(x+2)f(x)+2(x+2{)}^2{f}^{\prime }(x)-6(x+2)=10(x+2)f(x)$

Divide through by $2(x+2)$ (valid for $x\ge 0$):

$(x+2)f^{\prime }(x)-3f(x)=3$

This is a linear ODE in $y=f(x)$:

$\frac{dy}{dx}-\frac{3}{x+2}y=\frac{3}{x+2}$

Integrating factor $\mu =(x+2{)}^{-3}$. Multiply and integrate:

$\frac{d}{dx}\left[\frac{y}{(x+2{)}^3}\right]=\frac{3}{(x+2{)}^4}$

$\frac{y}{(x+2{)}^3}=-\frac{1}{(x+2{)}^3}+C\Rightarrow y=-1+C(x+2{)}^3$

Find $C$ from the original integral equation at $x=0$ (RHS integral $=0$):

$2(2{)}^{2}f(0)-3(2{)}^2=0\Rightarrow 8f(0)=12\Rightarrow f(0)=\frac{3}{2}$

$\frac{3}{2}=-1+8C\Rightarrow C=\frac{5}{16}$

$f(2)=-1+\frac{5}{16}\cdot 64=-1+20=19$.

$\therefore$ Answer: $19$.