The general solution of the differential equa | vedclass

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

Vedclass · Accepted Answer

$y = cx^{-2} + \frac{x^3}{5}$

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(D) The given differential equation is $\frac{dy}{dx} + \frac{2}{x}y = x^2$.This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \frac{2}{x}$ and $Q = x^2$.First,we calculate the integrating factor ($I$.$F$.):$I.F. = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = e^{\ln(x^2)} = x^2$.The general solution is given by $y \cdot (I.F.) = \int (Q \cdot I.F.) dx + c$.Substituting the values,we get $y \cdot x^2 = \int (x^2 \cdot x^2) dx + c$.$y \cdot x^2 = \int x^4 dx + c$.$y \cdot x^2 = \frac{x^5}{5} + c$.Dividing both sides by $x^2$,we get $y = \frac{x^3}{5} + cx^{-2}$.

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

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