Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}& f(x)=x-1\\ & f(f(x))=f(x)-1=x-1-1=x-2\\ & g(f(f(x)))=e^{x-2}\\ & \therefore \frac{dy}{dx}=e^{-2\sqrt{x}}\times e^{x-2}-\frac{1}{\sqrt{x}}y\\ & \frac{dy}{dx}+\frac{1}{\sqrt{x}}y=e^{x-2\sqrt{x}-2}\text{ which is L.D.E }\\ & \text{ I.F. }=e^{\int \frac{dy}{\sqrt{x}}}=e^{2\sqrt{x}}\end{array}$
Its solution is
$\begin{array}{rl}& y\times e^{2\sqrt{x}}=\int e^{2\sqrt{x}}\times e^{x-2\sqrt{x}-2}dx+c\\ & y\times e^{2\sqrt{x}}=\int e^{x-2}dx+c\\ & y\times e^{2\sqrt{x}}=e^{x-2}+c\end{array}$
$\text{ Given }\mathrm{x}=0,\mathrm{y}=0\Rightarrow 0={\mathrm{e}}^{-2}+\mathrm{c};\mathrm{c}=-{\mathrm{e}}^{-2}$
$\therefore \mathrm{y}\times {\mathrm{e}}^{2\sqrt{\mathrm{x}}}={\mathrm{e}}^{\mathrm{x}-2}-{\mathrm{e}}^{-2}$
$\text{ when }\mathrm{x}=1,\mathrm{y}\times {\mathrm{e}}^2={\mathrm{e}}^{-1}-{\mathrm{e}}^{-2}$
$y=\frac{e^{-1}-e^{-2}}{e^2}=\frac{\frac{1}{e}-\frac{1}{e^2}}{e^2}=\frac{e^2-e}{e^5}=\frac{e-1}{e^4}$
Option (1) is correct