Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}& {\mathrm{f}}^{\prime \prime }(\mathrm{x})=\mathrm{f}(\mathrm{x})\\ & \Rightarrow {\mathrm{f}}^{\prime }(\mathrm{x})\cdot {\mathrm{f}}^{\prime \prime }(\mathrm{x})={\mathrm{f}}^{\prime }(\mathrm{x})\cdot \mathrm{f}(\mathrm{x})\\ & \Rightarrow \frac{{\left({\mathrm{f}}^{\prime }(\mathrm{x})\right)}^2}{2}=\frac{(\mathrm{f}(\mathrm{x}){)}^2}{2}+\mathrm{C}\\ & \Rightarrow {\left({\mathrm{f}}^{\prime }(\mathrm{x})\right)}^2=(\mathrm{f}(\mathrm{x}){)}^2+{\mathrm{C}}^{\prime }\end{array}$
$\mathrm{f}(0)=0,{\mathrm{f}}^{\prime }(0)=3\Rightarrow {\mathrm{C}}^{\prime }=9$
$\therefore {\left({\mathrm{f}}^{\prime }(\mathrm{x})\right)}^2=(\mathrm{f}(\mathrm{x}){)}^2+9$
${\mathrm{f}}^{\prime }(\mathrm{x})=\sqrt{(\mathrm{f}(\mathrm{x}){)}^2+9}\because {\mathrm{f}}^{\prime }(\mathrm{x})\ge 0$
$\int \frac{\mathrm{dy}}{\sqrt{{\mathrm{y}}^2+9}}=\int \mathrm{dx}\Rightarrow \ln \left|\mathrm{y}+\sqrt{{\mathrm{y}}^2+9}\right|=\mathrm{x}+\mathrm{C}$
$\Rightarrow \mathrm{f}(0)=0\Rightarrow \mathrm{C}=\ln 3$
$\Rightarrow \mathrm{y}+\sqrt{{\mathrm{y}}^2+9}=3{\mathrm{e}}^{\mathrm{x}}$
at $x=\ln 3;y=4$
$\therefore 9\mathrm{f}(\ln 3)=36$