The general solution of the differential equa | vedclass

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(C) The given differential equation is $\frac{dy}{dx} + \frac{1}{x}y = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + Py =

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$xy = \frac{x^4}{4} + C$

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(C) The given differential equation is $\frac{dy}{dx} + \frac{1}{x}y = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = \frac{1}{x}$ and $Q = x^2$. First, we find the integrating factor $(IF)$: $IF = e^{\int P dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. The general solution is given by $y \cdot IF = \int (Q \cdot IF) dx + C$. Substituting the values, we get $y \cdot x = \int (x^2 \cdot x) dx + C$. $xy = \int x^3 dx + C$. Integrating $x^3$, we get $xy = \frac{x^4}{4} + C$.

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

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