Let f(x)=x-1 and g(x)=e^x for x R.

Question

Let $f(x)=x-1$ and $g(x)={e}^{x}$ for $x\in R$. If $\frac{dy}{dx}=({e}^{-2\sqrt{x}}g(f(f(x)))-\frac{y}{\sqrt{x}})$, $y(0)=0$, then $y(1)$ is

Accepted Answer

$$\begin{aligned}
& f(x)=x-1 \
& f(f(x))=f(x)-1=x-1-1=x-2 \
& g(f(f(x)))=e^{x-2} \
& \therefore \frac{d y}{d x}=e^{-2 \sqrt{x}} \times e^{x-2}-\frac{1}{\sqrt{x}} y \
& \frac{d y}{d x}+\frac{1}{\sqrt{x}} y=e^{x-2 \sqrt{x}-2} \text { which is L.D.E } \
& \text { I.F. }=e^{\int \frac{d y}{\sqrt{x}}}=e^{2 \sqrt{x}}
\end{aligned}$$

Its solution is

$$\begin{aligned}
& y \times e^{2 \sqrt{x}}=\int e^{2 \sqrt{x}} \times e^{x-2 \sqrt{x}-2} d x+c \
& y \times e^{2 \sqrt{x}}=\int e^{x-2} d x+c \
& y \times e^{2 \sqrt{x}}=e^{x-2}+c
\end{aligned}$$

Given $\mathrm{x}=0, \mathrm{y}=0 \Rightarrow 0=\mathrm{e}^{-2}+\mathrm{c} \quad ; \mathrm{c}=-\mathrm{e}^{-2}$

$$\therefore \mathrm{y} \times \mathrm{e}^{2 \sqrt{x}}=\mathrm{e}^{\mathrm{x}-2}-\mathrm{e}^{-2}$$

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

Solutions · 1