If y(x) is the solution of the differential e | vedclass

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(A) Given the differential equation: $(x + 2)\frac{dy}{dx} = x^2 + 4x - 9$.Rearranging for $dy$:$dy = \frac{x^2 + 4x - 9}{x + 2} dx$.Perform polynomia

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(A) Given the differential equation: $(x + 2)\frac{dy}{dx} = x^2 + 4x - 9$.Rearranging for $dy$:$dy = \frac{x^2 + 4x - 9}{x + 2} dx$.Perform polynomial division on the right side:$\frac{x^2 + 4x - 9}{x + 2} = \frac{x(x + 2) + 2x - 9}{x + 2} = x + \frac{2x - 9}{x + 2} = x + \frac{2(x + 2) - 13}{x + 2} = x + 2 - \frac{13}{x + 2}$.Integrating both sides:$y = \int (x + 2 - \frac{13}{x + 2}) dx = \frac{x^2}{2} + 2x - 13 \ln|x + 2| + C$.Using the initial condition $y(0) = 0$:$0 = \frac{0^2}{2} + 2(0) - 13 \ln|0 + 2| + C \implies C = 13 \ln 2$.So,the solution is $y(x) = \frac{x^2}{2} + 2x - 13 \ln|x + 2| + 13 \ln 2$.Now,find $y(-4)$:$y(-4) = \frac{(-4)^2}{2} + 2(-4) - 13 \ln|-4 + 2| + 13 \ln 2$$y(-4) = \frac{16}{2} - 8 - 13 \ln 2 + 13 \ln 2$$y(-4) = 8 - 8 = 0$.

If $y(x)$ is the solution of the differential equation $(x + 2)\frac{dy}{dx} = x^2 + 4x - 9, x \ne -2$ and $y(0) = 0$,then $y(-4)$ is equal to

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