The solution of the differential equation x^2 | vedclass

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Question

(C) Step $1$: Rewrite the differential equation in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.Divide by $x^2$: $\frac{dy}{dx} - \frac{1}{

Vedclass · Accepted Answer

$2xy - 2cx^2 + 1 = 0$

Vedclass · Answer

(C) Step $1$: Rewrite the differential equation in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$.Divide by $x^2$: $\frac{dy}{dx} - \frac{1}{x}y = \frac{1}{x^2}$.Step $2$: Find the integrating factor $IF = e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = \frac{1}{x}$.Step $3$: Multiply the equation by $IF$ and integrate: $\frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = \frac{1}{x^3}$.This is $\frac{d}{dx} (y \cdot \frac{1}{x}) = x^{-3}$.Step $4$: Integrate both sides: $y \cdot \frac{1}{x} = \int x^{-3} dx = \frac{x^{-2}}{-2} + c = -\frac{1}{2x^2} + c$.Step $5$: Multiply by $2x^2$: $2xy = -1 + 2cx^2$, which rearranges to $2xy - 2cx^2 + 1 = 0$.

The solution of the differential equation $x^2 \frac{dy}{dx} - xy = 1$ is...

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