The general solution of the differential equa | vedclass

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Question

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{x}$ and $Q = x^2 + 5$.

Vedclass · Accepted Answer

$xy = \frac{x^4}{4} + \frac{5x^2}{2} + c$

Vedclass · Answer

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{x}$ and $Q = x^2 + 5$.Step $1$: Find the integrating factor $(IF)$.$IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$.Step $2$: Multiply the differential equation by $IF$ and integrate.$x \frac{dy}{dx} + y = x(x^2 + 5) = x^3 + 5x$.$\frac{d}{dx}(xy) = x^3 + 5x$.Integrating both sides with respect to $x$:$xy = \int (x^3 + 5x) dx = \frac{x^4}{4} + \frac{5x^2}{2} + c$.

The general solution of the differential equation $\frac{dy}{dx} + \frac{y}{x} = x^2 + 5$ is ....

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