The integrating factor of the differential eq | vedclass

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Question

(A) The given differential equation is $x \frac{dy}{dx} + 2y = x^2 \log x$. Dividing both sides by $x$,we get: $\frac{dy}{dx} + \frac{2}{x} y = x \log

Vedclass · Accepted Answer

$x^2$

Vedclass · Answer

(A) The given differential equation is $x \frac{dy}{dx} + 2y = x^2 \log x$. Dividing both sides by $x$,we get: $\frac{dy}{dx} + \frac{2}{x} y = x \log x$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{2}{x}$ and $Q(x) = x \log x$. The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx}$. $IF = e^{\int \frac{2}{x} dx} = e^{2 \log x} = e^{\log x^2} = x^2$. Therefore,the correct option is $A$.

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

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