Question

If the solution of differential equation $\frac{dy}{dx}=y+{\int }_0^{1}ydx,$ is $y=\frac{1}{3-e}(A{e}^x+Be+1)$; also given that $y=1,$ when $x=0$,
then find the value of $A+B$.

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (B)

$1$

$\frac{dy}{dx}=y+{\int }_0^{1}ydx$
$\Rightarrow \frac{dy}{dx}=y+a$ let ${\int }_0^{1}ydx=a$
$\Rightarrow \frac{dy}{dx}-y-a=0$
I.F.$=e^{-\int 1.dx}=e^{-x}$
Hence, the solution of the given first order linear differential equation is
$y\times e^{-x}=\int a{e}^{-x}dx+c$
$\Rightarrow y=-a+c{e}^x....\left(1\right)$
$a={\int }_0^{1}ydx=-{\int }_0^1(a-c{e}^x)dx$
$=-a(1-0)+{\left|c\right(e^x\left)\right|}_0^1=-a+c(e-1)$
$\Rightarrow a=\frac{(e-1)c}{2}...\left(2\right)$
Using $\left(1\right)$ and $\left(2\right)$ we get,
$y=\frac{(1-e)c}{2}+c{e}^x$
Given that $y=1$, when $x=0$. Using this we get
$1=\frac{(1-e)c}{2}+c\Rightarrow 2=c(3-e)\Rightarrow c=\frac{2}{3-e}$
Hence $y=\frac{(1-e)c}{2}+c{e}^{x}y=c{e}^x-\frac{c}{2}e+\frac{c}{2}y=\frac{1}{3-e}({2e}^x-e+1)$
Comparing this with given value of $y$, we get $A+B=1$