The solution of frac{dy}{dx} + p(x)y = 0 is | vedclass

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(C) The given equation is $\frac{dy}{dx} + p(x)y = 0$. This is a first-order linear differential equation. We can solve this by separating the variabl

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$y = c e^{-\int p dx}$

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(C) The given equation is $\frac{dy}{dx} + p(x)y = 0$. This is a first-order linear differential equation. We can solve this by separating the variables: $\frac{dy}{y} = -p(x) dx$. Integrating both sides,we get: $\int \frac{dy}{y} = -\int p(x) dx$. $\ln|y| = -\int p(x) dx + C_1$. Taking the exponential of both sides: $|y| = e^{-\int p(x) dx + C_1} = e^{C_1} e^{-\int p(x) dx}$. Let $e^{C_1} = c$,then the solution is $y = c e^{-\int p(x) dx}$.

The solution of $\frac{dy}{dx} + p(x)y = 0$ is

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