Let , $y(x)$ be a continuous solution of the initial value problem $y’+2y=f(x)$ , $y(0)=0$ , where, $$f(x)=\begin{cases}1 & \text{ if } 0\le x\le 1\\0 & \text{ if } x>1\end{cases}$$Then, $y(3/2)$ equals to
(A) $\frac{\sinh(1)}{e^3}$
(B) $\frac{\cosh(1)}{e^3}$
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(C) $\frac{\sinh(1)}{e^2}$
(D) $\frac{\cosh(1)}{e^2}$
Integrating both sides we get , $$\int_0^xd(y(x))+2\int_0^xy(x)\,dx=\int_0^xf(x)\,dx$$
$$\implies y(x)+2\int_0^xy(x)\,dx=\int_0^xf(x)\,dx=1$$
$$\implies y(x)=1-2C$$where , $$C=\int_0^xy(x)\,dx=(1-2C)x\implies C =\frac{x}{1+2x}$$
Then , $y(x)=\frac{1}{1+2x}$ and so, $y(3/2)=1/4$.
But no option match..Please help..
First of all multiply both sides by the integrating factor $e^{2x}$ to get
$$(e^{2x}y(x))’=e^{2x}f(x) $$
Integrating both sides from $0$ to $3/2$ gives
$$e^3y(3/2)-e^{0}y(0)=\int_0^{3/2} e^{2x}f(x) \mathrm{d} x.$$
Using the initial data we get
$$y(3/2)=e^{-3} \int_0^{3/2} e^{2x}f(x) \mathrm{d}x $$
and I leave the evaluation of the integral to you.
EDIT:
As for what’s wrong with your attempt:
The integral $\int_0^x f(x) \mathrm{d}x$ is not always 1. This is true only if $x \geq 1$ which is unclear to me.
What you denoted by $C$ is not a constant actually.
`Solving the Part (1)
$y′+2y=1 , y(0)=0$
We get
$$y(x) = \frac{(1-e^{-2x}) }{2}…………………………………….(A)$$
Consider the Part(2),
$y′+2y=0 $
$y(x)=Ke^{-2x}$
$y(1)=1/2*(1-e^{-2})$ By the continuity find $y(1)$ from equation $(A)$
$ K=y(1)e^2$
hence
$y(x)=y(1)e^2e^{-2x}$
$y(3/2)=sinh(1)/(e^2)$
$So$ $Option $ (C) $ is$ $ the$ $ correct $ $Answer$