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Q.
If $\frac{d y}{d x}=y+\int \limits_{0}^{1} y \,d x,$ given $y=1,$ where $x=0,$ then the
value of $y(1)$ is
Differential Equations
Solution:
$\frac{d y}{d x}=y+\int \limits_{0}^{1} y \,d x \,\,\,\,\,\,\, (i)$
$\Rightarrow \frac{d y}{d x}=y+k_{1} \,\,\,\,\,\,\, $ $\left(\right.$ let $\left.k_{1}=\int \limits_{0}^{1} y d x\right)$
$\Rightarrow \int \frac{d y}{y+k_{1}}=\int d x $
$\Rightarrow \log _{e}\left(y+k_{1}\right)=x+c$
$\Rightarrow y+k_{1}=k_{2} e^{x} \,\,\,\,\,\,\, (ii)$
Curve passes through $(0,1) \Rightarrow 1+k_{1}=k_{2} \,\,\,\,\,\,\,(iii)$
Putting the value of $y$ (from equation(ii) into equation (i))
$ \frac{d y}{d x}=\left(k_{2} e^{x}-k_{1}\right)+\int \limits_{0}^{1}\left(k_{2} e^{x}-k_{1}\right) d x $
$\Rightarrow \frac{d y}{d x}=\left(k_{2} e^{x}-k_{1}\right)+\left[k_{2} e^{x}-k_{1} x\right]_{0}^{1} $
$\Rightarrow \frac{d y}{d x}=k_{2} e^{x}-k_{1}+k_{2} e-k_{1}-k_{2} $
$\Rightarrow d y=\left(k_{2} e^{x}+k_{2} e-2 k_{1}-k_{2}\right) d x$
$\Rightarrow \int \limits_{1}^{y} d y=\int \limits_{0}^{x}\left(k_{2} e^{x}+k_{2} e-2 k_{1}-k_{2}\right) d x $
$\Rightarrow y-1=k_{2} e^{x}+\left(k_{2} e-2 k_{1}-k_{2}\right) x-k_{2} \,\,\,\,\,\,\,\, (iv)$
Since equations (ii) and (iv) are similar
$\therefore k_{2} e-2 k_{1}-k_{2}=0$ 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} $
$\therefore y =\frac{1}{3-e}\left(2 e^{x}-e+1\right)$