The solution of the differential equation x f | vedclass

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Question

(B) Given the differential equation: $x \frac{dy}{dx} + 2y = x^2$.Divide by $x$ to get the standard form of a linear differential equation: $\frac{dy}

Vedclass · Accepted Answer

$y = \frac{x^2}{4} + \frac{3}{4x^2}$

Vedclass · Answer

(B) Given the differential equation: $x \frac{dy}{dx} + 2y = x^2$.Divide by $x$ to get the standard form of a linear differential equation: $\frac{dy}{dx} + \frac{2}{x}y = x$.Here,$P(x) = \frac{2}{x}$ and $Q(x) = x$.The Integrating Factor $(IF)$ is given by: $IF = e^{\int P(x) dx} = e^{\int \frac{2}{x} dx} = e^{2 \ln|x|} = x^2$.The general solution is: $y \cdot IF = \int Q(x) \cdot IF dx + C$.$y \cdot x^2 = \int x \cdot x^2 dx = \int x^3 dx = \frac{x^4}{4} + C$.Using the condition $y(1) = 1$:$1 \cdot (1)^2 = \frac{1^4}{4} + C \Rightarrow 1 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4}$.Substituting $C$ back into the general solution:$y x^2 = \frac{x^4}{4} + \frac{3}{4}$.Dividing by $x^2$,we get: $y = \frac{x^2}{4} + \frac{3}{4x^2}$.

The solution of the differential equation $x \frac{dy}{dx} + 2y = x^2$ $(x \neq 0)$ with $y(1) = 1$ is

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