The general solution of the differential equa | vedclass

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Question

(A) Given the differential equation: $\frac{dy}{dx} + xy = 4x - 2y + 8$. Rearranging the terms to standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} + xy = 4x - 2y + 8$. Rearranging the terms to standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$: $\frac{dy}{dx} + (x + 2)y = 4x + 8$. $\frac{dy}{dx} + (x + 2)y = 4(x + 2)$. Here,$P(x) = x + 2$ and $Q(x) = 4(x + 2)$. The integrating factor $IF = e^{\int P(x) dx} = e^{\int (x + 2) dx} = e^{\frac{x^2}{2} + 2x}$. The general solution is given by $y \cdot IF = \int Q(x) \cdot IF dx + c$. $y \cdot e^{\frac{x^2}{2} + 2x} = \int 4(x + 2) e^{\frac{x^2}{2} + 2x} dx + c$. Let $u = \frac{x^2}{2} + 2x$,then $du = (x + 2) dx$. $y \cdot e^{\frac{x^2}{2} + 2x} = 4 \int e^u du + c = 4e^u + c = 4e^{\frac{x^2}{2} + 2x} + c$. Dividing both sides by $e^{\frac{x^2}{2} + 2x}$: $y = 4 + ce^{-(\frac{x^2}{2} + 2x)}$. Thus,the correct option is $A$.

The general solution of the differential equation $\frac{dy}{dx} + xy = 4x - 2y + 8$ is

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