JEE Main 2025 Apr 2 Shift 1, Mathematics Q22: Methods of Solving First Order, First Degree Differential Equation

Multiply $f^{\prime \prime }(x)=f(x)$ by $f^{\prime }(x)$ and integrate:

$$f^{\prime }(x)f^{\prime \prime }(x)=f(x)f^{\prime }(x)\Rightarrow \frac{d}{dx}\left[\frac{(f^{\prime }(x){)}^2}{2}\right]=\frac{d}{dx}\left[\frac{(f(x){)}^2}{2}\right]$$

$$\Rightarrow (f^{\prime }(x){)}^2=(f(x){)}^2+C$$

Use $f(0)=0$, $f^{\prime }(0)=3$: $9=0+C$, so $C=9$.

$$f^{\prime }(x)=\sqrt{f(x{)}^2+9}\text{(positive root, since }{f}^{\prime }\ge 0\text{)}$$

Separate variables with $y=f(x)$:

$$\int \frac{dy}{\sqrt{y^2+9}}=\int dx\Rightarrow \ln \left|y+\sqrt{y^2+9}\right|=x+C_1$$

At $x=0$, $y=0$: $\ln 3=C_1$.

$$\Rightarrow y+\sqrt{y^2+9}=3e^x$$

At $x=\ln 3$: $y+\sqrt{y^2+9}=9$, so $\sqrt{y^2+9}=9-y$. Squaring:

$$y^2+9=81-18y+y^2\Rightarrow 18y=72\Rightarrow y=4$$

Hence $f(\ln 3)=4$ and $9f(\ln 3)=36$.