Question

If $\frac{dy}{dx}=y+\underset{0}{\overset{1}{\int }}ydx,$ given $y=1,$ where $x=0,$ then the value of $y(1)$ is

• A. $\frac{e+1}{3-e}$
• B. $\frac{2e+1}{3-e}$
• C. $\frac{e+1}{3-2e}$
• D. $\frac{2e+1}{3-2e}$

Answer 1

Answer: (A)

$\frac{dy}{dx}=y+\underset{0}{\overset{1}{\int }}ydx(i)$
$\Rightarrow \frac{dy}{dx}=y+k_1$ $($ let $k_1=\underset{0}{\overset{1}{\int }}ydx)$
$\Rightarrow \int \frac{dy}{y+k_1}=\int dx$
$\Rightarrow {\log }_e\left(y+k_1\right)=x+c$
$\Rightarrow y+k_1=k_2{e}^x(ii)$
Curve passes through $(0,1)\Rightarrow 1+k_1=k_2(iii)$
Putting the value of $y$ (from equation(ii) into equation (i))
$\frac{dy}{dx}=\left(k_2{e}^x-k_1\right)+\underset{0}{\overset{1}{\int }}\left(k_2{e}^x-k_1\right)dx$
$\Rightarrow \frac{dy}{dx}=\left(k_2{e}^x-k_1\right)+{\left[k_2{e}^x-k_{1}x\right]}_0^1$
$\Rightarrow \frac{dy}{dx}=k_2{e}^x-k_1+k_{2}e-k_1-k_2$
$\Rightarrow dy=\left(k_2{e}^x+k_{2}e-2k_1-k_2\right)dx$
$\Rightarrow \underset{1}{\overset{y}{\int }}dy=\underset{0}{\overset{x}{\int }}\left(k_2{e}^x+k_{2}e-2k_1-k_2\right)dx$
$\Rightarrow y-1=k_2{e}^x+\left(k_{2}e-2k_1-k_2\right)x-k_2(iv)$
Since equations (ii) and (iv) are similar
$\therefore k_{2}e-2k_1-k_2=0$ and $1+k_1=k_2$
Solving equations, we get
$k_2=\frac{2}{3-e},k_1=\frac{e-1}{3-e}$
From equation (ii), $y+k_1=k_2{e}^x$
$\Rightarrow y=\frac{2}{3-e}{e}^x-\frac{e-1}{3-e}$
$\therefore y=\frac{1}{3-e}\left(2e^x-e+1\right)$