Solve the initial value problem: x d y d x + y = x x , y (1)=1 4 - Vidyadip

Question

Solve the initial value problem: $x \frac{d y}{d x}+ y = x \log x , y (1)=\frac{1}{4}$

✓

Answer

We have, $x \frac{d y}{d x}+ y = x \log x$
$
\Rightarrow \frac{d y}{d x}+\frac{1}{x} y=\log x \ldots(i)
$
This is linear differential equation of the form $\frac{d y}{d x}+ Py = Q$ with $P =\frac{1}{x}$ and $Q =\log x$
$
\therefore \text { I.F. }=e^{\int \frac{1}{x} d x}=e^{\log x}=x[\because x>0]
$
Multiplying both sides of (i) by I.F. $=x$, we get
$
x \frac{d y}{d x}+y=x \log x
$
Integrating with respect to $x$, we ge

It is given that $y(1)=\frac{1}{4}$ i.e. $y=\frac{1}{4}$ where $x=1$. Putting $x=1$ and $y=\frac{1}{4}$ in (ii), we get
$
\frac{1}{4}=0-\frac{1}{4}+C \Rightarrow C=\frac{1}{2}
$
Putting $C=\frac{1}{2}$ in (ii), we get
$
xy=\frac{x^2}{2}(\log x)-\frac{x^2}{2}+\frac{1}{2} \Rightarrow y=\frac{1}{2} x \log x-\frac{x}{4}+\frac{1}{2 x}
$
Hence, $y =\frac{1}{2} x \log x -\frac{x}{4}+\frac{1}{2 x}$ is the solution of the given differential equation.

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