The general solution of the differential equa | vedclass

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Question

(B) The given differential equation is $\frac{dy}{dx} + g'(x)y = g(x)g'(x)$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y

Vedclass · Accepted Answer

$g(x) + \log|y - g(x) + 1| = c$

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(B) The given differential equation is $\frac{dy}{dx} + g'(x)y = g(x)g'(x)$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = g'(x)$ and $Q(x) = g(x)g'(x)$.The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx} = e^{\int g'(x) dx} = e^{g(x)}$.The general solution is $y(IF) = \int Q(IF) dx + C$.Substituting the values,we get $y e^{g(x)} = \int g(x)g'(x) e^{g(x)} dx + C$.Let $g(x) = t$,then $g'(x) dx = dt$.So,$y e^{g(x)} = \int t e^t dt + C$.Using integration by parts,$\int t e^t dt = t e^t - e^t + C = e^t(t - 1) + C$.Thus,$y e^{g(x)} = e^{g(x)}(g(x) - 1) + C$.Rearranging the terms,$y e^{g(x)} - e^{g(x)}(g(x) - 1) = C$.$e^{g(x)}(y - g(x) + 1) = C$.Taking the natural logarithm on both sides,$\log(e^{g(x)}) + \log|y - g(x) + 1| = \log(C)$.$g(x) + \log|y - g(x) + 1| = C'$.Thus,option $(b)$ is correct.

The general solution of the differential equation $\frac{dy}{dx} + y g'(x) = g(x) g'(x)$ is

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