Question

Solve:

$\frac{dy}{dx}=y+{\int }_0^{1}ydx$ given $y=1$, where $x=0$

• A. $y=\frac{1}{4+e}(2e^x+e+2)$
• B. $y=\frac{1}{3+e}(2e^x+e+1)$
• C. $y=\frac{1}{2-e}(e^x-e+1)$
• D. $y=\frac{1}{3-e}(2e^x-e+1)$

Answer 1

Answer: (D)

$y=\frac{1}{3-e}(2e^x-e+1)$

$\frac{dy}{dx}=y+{\int }_0^{1}ydx$
Clearly ${\int }_0^{1}ydx=$ constant $=k$ (say)
So differential equation becomes, $\frac{dy}{dx}=y+k\Rightarrow \frac{dy}{y+k}=dx$
$\Rightarrow \log (y+k)=x+c$, $c$ is integration constant.
Give at $x=0,y=1\Rightarrow c=\log (1+k)\therefore y=e^x{e}^c-k=e^x(1+k)-k$
Now $k={\int }_0^1\left[\right(1+k)e^x-k]dx=(e-1)(1+k)-k\Rightarrow k=\frac{e-1}{3-e}$
substituting in above equation we get,
$y=\frac{1}{3-e}(2e^x-e+1)$