Let x=x(y) be the solution of the differential equation y^2 d x+ … - Vidyadip

MCQ

Let $x=x(y)$ be the solution of the differential equation $y^2 d x+\left(x-\frac{1}{y}\right) d y=0$. If $x(1)=1$, then $x\left(\frac{1}{2}\right)$ is :

A
$\frac{1}{2}+ e$
B
$\frac{1}{2}+ e$
✓
$3- e$
D
$3+e$

✓

Answer

Correct option: C.

$3- e$

(C) $3- e$
Sol.
$
\begin{array}{l}
\frac{d x}{d y}+\left(\frac{1}{y^2}\right) x=\frac{1}{y^3} \\
\text { I.F. }=e^{\int -\frac{1}{y^2} d y}=e^{-\frac{1}{y}} \\
\Rightarrow x \cdot e^{-\frac{1}{y}}=\int\left(e^{-\frac{1}{t}}\right) \cdot \frac{1}{y^3} d y \\
\text { Put }-\frac{1}{y}=t \\
+\frac{1}{y^2} d y=d t \\
x . e^{-\frac{1}{y}}=-\int t \cdot e^t d t \\
x . e^{-\frac{1}{y}}=-t e^t+e^t+C \\
x . e^{-\frac{1}{y}}=\frac{+1}{y} e^{-\frac{1}{y}}+e^{-\frac{1}{y}}+C \\
x=1, y=1
\end{array}$
$\begin{array}{l}\frac{1}{ e }=\frac{1}{ e }+\frac{1}{ e }+ C \\ \Rightarrow C =-\frac{1}{ e } \\ \text { Put } y =\frac{1}{2} \\ \frac{ x }{ e ^2}=\frac{2}{ e ^2}+\frac{1}{ e ^2}-\frac{1}{ e } \\ x =3- e \end{array}$

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