The general solution of the differential equa | vedclass

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Question

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

Vedclass · Accepted Answer

$y = ce^{-\phi(x)} + \phi(x) - 1$

Vedclass · Answer

(A) The given differential equation is $\frac{dy}{dx} + y\phi'(x) = \phi(x)\phi'(x)$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \phi'(x)$ and $Q = \phi(x)\phi'(x)$.The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{\int \phi'(x) dx} = e^{\phi(x)}$.The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$.$y \cdot e^{\phi(x)} = \int \phi(x)\phi'(x) e^{\phi(x)} dx + c$.Let $t = \phi(x)$,then $dt = \phi'(x) dx$.The integral becomes $\int t e^t dt = t e^t - e^t + c$.Substituting back,$y \cdot e^{\phi(x)} = \phi(x)e^{\phi(x)} - e^{\phi(x)} + c$.Dividing both sides by $e^{\phi(x)}$,we get $y = \phi(x) - 1 + c e^{-\phi(x)}$.

The general solution of the differential equation,$y' + y\phi'(x) - \phi(x)\phi'(x) = 0$,where $\phi(x)$ is a known function,is: (where $c$ is an arbitrary constant)

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