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y^(')+2(y)/(x-3)=(sin 4x)/((x-3))

$y^{\prime}+2 \frac{y}{x-3}=\frac{\sin 4 x}{(x-3)}$

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Find derivatives using the chain rule I

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Q.

y^{\prime}+2 \frac{y}{x-3}=\frac{\sin 4 x}{(x-3)}

Identify Differential Equation: Identify the given differential equation. The equation is $y' + \frac{2y}{x-3} = \frac{\sin 4x}{x-3}$ .
Recognize Linear Form: Notice that the equation is a first-order linear differential equation in the form $y' + P(x)y = Q(x)$ , where $P(x) = \frac{2}{x-3}$ and $Q(x) = \frac{\sin 4x}{x-3}$ .
Find Integrating Factor: To solve this equation, we need to find an integrating factor, which is $e^{(\int P(x)\,dx)}$ . Calculate the integral of $P(x)$ to find the integrating factor.
Integrate $P(x)$ : Integrate $P(x) = \frac{2}{x-3}$ . The integral of $\frac{2}{x-3}$ with respect to $x$ is $2\ln|x-3|$ .
Multiply by Integrating Factor: The integrating factor is $e^{2\ln|x-3|}$ , which simplifies to $(x-3)^2$ .
Recognize Derivative Form: Multiply the entire differential equation by the integrating factor $(x-3)^2$ to get $(x-3)^2y' + 2(x-3)y = \sin 4x$ .
Integrate Both Sides: Recognize that the left side of the equation is now the derivative of $(x-3)^2y$ with respect to $x$ . Write the equation as $\frac{d}{dx}[(x-3)^2y] = \sin 4x$ .
Final Integration: Integrate both sides of the equation with respect to $x$ . The integral of $\frac{d}{dx}[(x-3)^2y]$ is $(x-3)^2y$ , and the integral of $\sin 4x$ is $-\frac{1}{4} \cos 4x$ plus a constant $C$ .
Solve for y: After integrating, we have $(x-3)^2y = -\frac{1}{4} \cos 4x + C$ .
Solve for y: After integrating, we have $(x-3)^2y = -\frac{1}{4} \cos 4x + C$ . Solve for y by dividing both sides by $(x-3)^2$ . The solution for y is $y = \frac{-\frac{1}{4} \cos 4x + C}{(x-3)^2}$ .

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