Solve the following differential equations : x d y d x +2 y=x^2 x - Vidyadip

Question

Solve the following differential equations : $x \frac{d y}{d x}+2 y=x^2 \cdot \log x$

✓

Answer

$
\begin{aligned}
& x \frac{d y}{d x}+2 y = x ^2 \cdot \log x \\
& \therefore \frac{d y}{d x}+\left(\frac{2}{x}\right) \cdot y=x \cdot \log x
\end{aligned}
$
This is the linear differential equation of the form
$
\begin{aligned}
& \frac{d y}{d x}+P \cdot y=Q \text {, where } P=\frac{2}{x} \text { and } Q=x \cdot \log x \\
& \therefore \text { I.F. }=e^{\int P d x}=e^{\int \frac{2}{x} d x}=e^{2 \int \frac{1}{x} d x} \\
& =e^{2 \log x}=e^{\log x^2}=x^2
\end{aligned}
$
$\therefore$ the solution of (1) is given by
$
\begin{aligned}
y \cdot(\text { I.F. }) & =\int Q \cdot \text { (I.F.) } d x+c \\
\therefore y \cdot x^2 & =\int(x \log x) \cdot x^2 d x+c \\
\therefore x^2 \cdot y & =\int x^3 \cdot \log x d x+c \\
= & (\log x) \int x^3 d x-\int\left[\frac{d}{d x}(\log x) \int x^3 d x\right] d x+c \\
= & (\log x) \cdot \frac{x^4}{4}-\int \frac{1}{x} \cdot \frac{x^4}{4} d x+c \\
= & \frac{1}{4} x^4 \log x-\frac{1}{4} \int x^3 d x+c
\end{aligned}
$
$
\begin{aligned}
& \therefore x^2 \cdot y=\frac{1}{4} x^4 \log x-\frac{1}{4} \cdot \frac{x^4}{4}+c \\
& \therefore y \cdot x^2=\frac{x^4 \log x}{4}-\frac{x^4}{16}+c
\end{aligned}
$
This is the general solution.

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