The general solution of frac{dy}{dx} + y f^{p | vedclass

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(C) The given differential equation is $\frac{dy}{dx} + y f^{\prime}(x) = f(x) f^{\prime}(x)$.This is a linear differential equation of the form $\fra

Vedclass · Accepted Answer

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

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(C) The given differential equation is $\frac{dy}{dx} + y f^{\prime}(x) = f(x) f^{\prime}(x)$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = f^{\prime}(x)$ and $Q(x) = f(x) f^{\prime}(x)$.The integrating factor $IF = e^{\int P(x) dx} = e^{\int f^{\prime}(x) dx} = e^{f(x)}$.The general solution is given by $y \cdot IF = \int Q(x) \cdot IF dx + C$.Substituting the values, we get $y \cdot e^{f(x)} = \int f(x) f^{\prime}(x) e^{f(x)} dx + C$.Let $u = f(x)$, then $du = f^{\prime}(x) dx$. The integral becomes $\int u e^u du$.Using integration by parts, $\int u e^u du = u e^u - e^u$.So, $y \cdot e^{f(x)} = f(x) e^{f(x)} - e^{f(x)} + C$.Dividing by $e^{f(x)}$, we get $y = f(x) - 1 + C e^{-f(x)}$.

The general solution of $\frac{dy}{dx} + y f^{\prime}(x) - f(x) f^{\prime}(x) = 0$, where $y \neq f(x)$, is

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