Question

The solution of the differential equation $x\frac{dy}{dx}+2y=x^2\log x$ is

• A. $\frac{x^2}{16}(4\log x-1)+C{x}^{-2}$
• B. $\frac{x^2}{16}(2\log x-1)+C{x}^{-2}$
• C. $\frac{x^2}{4}(4\log x-1)+C{x}^{-2}$
• D. None of the above

Answer 1

Answer: (A)

$\frac{x^2}{16}(4\log x-1)+C{x}^{-2}$

Given differential equation can be written as
$\frac{dy}{dx}+\frac{2y}{x}=x\log x$
Here, $P=\frac{2}{x}$ and $Q=x\log x$
$\therefore IF=e^{\int Pdx}=e^{2\int \frac{1}{x}dx}$
$=e^{2\log x}=e^{\log x^2}=x^2$
$\therefore$ The general solution is
$y\times IF=\int Q\times IF$
$\Rightarrow y\times x^2=\int \times \log x\times x^{2}dx+C$
$=\int x^3\log xdx$
$=\log x\times \frac{x^4}{4}-\int [\frac{1}{x}\times \frac{x^4}{4}]dx+C$
$=\frac{x^4}{4}\log x-\int \frac{x^3}{4}dx+C$
$\Rightarrow x^{2}y=\frac{x^4}{4}\log x-\frac{x^4}{16}+C$
$\Rightarrow y=\frac{x^2}{16}(4\log x-1)+C{x}^{-2}$.