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Question

(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x)

Vedclass · Accepted Answer

$2y = cx^2 + x^4$

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(C) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{2}{x}$ and $Q(x) = x^3$.First,we find the integrating factor $(IF)$:$IF = e^{\int P(x) dx} = e^{\int -\frac{2}{x} dx} = e^{-2 \ln|x|} = e^{\ln|x^{-2}|} = x^{-2} = \frac{1}{x^2}$.The general solution is given by $y 	imes (IF) = \int Q(x) 	imes (IF) dx + c$.Substituting the values:$y 	imes \frac{1}{x^2} = \int x^3 	imes \frac{1}{x^2} dx + c$$y 	imes \frac{1}{x^2} = \int x dx + c$$y 	imes \frac{1}{x^2} = \frac{x^2}{2} + c$Multiplying both sides by $x^2$:$y = \frac{x^4}{2} + cx^2$$2y = x^4 + 2cx^2$.Since $2c$ is an arbitrary constant,we can write it as $C$.Thus,$2y = x^4 + Cx^2$.

The solution of the differential equation $\frac{dy}{dx} - 2\frac{y}{x} = x^3$ is:

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