Find the general solution of the differential equation,
$x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$

Solution

$x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$

Dividing all the terms of the equation by x log x, we get

$\Rightarrow \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x^{2}}$

This equation is in the form of a linear differential equation

$\frac{dy}{dx} + py = Q, where P = \frac{1}{x \log x} and Q = \frac{2}{x^{2}} Now, I . F = e^{\int pdx} = e^{\int \frac{1}{x \log x} dx} = e^{\log (\log x)} = \log x$

The general solution of the given differential equation is given by

y x I.F. = $\int$ ( Q x I.F. ) dx + C

$\Rightarrow y \log x = \int (\frac{2}{x^{2}} \log x) dx \Rightarrow y \log x = 2 \int (\log x \times \frac{1}{x^{2}}) dx . = 2 [\log x \times \int \frac{1}{x^{2}} dx - \int \{\frac{d}{dx} (\log x) \times \int \frac{1}{x^{2}} dx\} dx] = 2 [\log x (- \frac{1}{x}) - \int (\frac{1}{x} \times (- \frac{1}{x})) dx] = 2 [- \frac{\log x}{x} + \int \frac{1}{x^{2}} dx] = 2 [- \frac{\log x}{x} - \frac{1}{x}] + C So the required general solution is y \log x = - \frac{2}{x} (1 + \log x) + C$

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Differential Equations

Find the general solution of the differential equation,
$x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$

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For Daily Free Study Material Join wiredfaculty

Differential Equations

Find the general solution of the differential equation,x log x dydx + y = 2x log x

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the general solution of the differential equation,
$x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$