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

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(A) Given the differential equation: $(x+2) \frac{dy}{dx} = x^2+4x-9$.We can rewrite the right side as: $x^2+4x-9 = (x^2+4x+4) - 13 = (x+2)^2 - 13$.So

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(A) Given the differential equation: $(x+2) \frac{dy}{dx} = x^2+4x-9$.We can rewrite the right side as: $x^2+4x-9 = (x^2+4x+4) - 13 = (x+2)^2 - 13$.So,$\frac{dy}{dx} = \frac{(x+2)^2 - 13}{x+2} = (x+2) - \frac{13}{x+2}$.Integrating both sides with respect to $x$:$\int dy = \int (x+2) dx - 13 \int \frac{1}{x+2} dx$.$y = \frac{(x+2)^2}{2} - 13 \ln|x+2| + C$.Given $y(0) = 0$,substitute $x=0$ and $y=0$:$0 = \frac{(0+2)^2}{2} - 13 \ln|0+2| + C$.$0 = 2 - 13 \ln(2) + C \implies C = 13 \ln(2) - 2$.Thus,the solution is $y(x) = \frac{(x+2)^2}{2} - 13 \ln|x+2| + 13 \ln(2) - 2$.Now,find $y(-4)$:$y(-4) = \frac{(-4+2)^2}{2} - 13 \ln|-4+2| + 13 \ln(2) - 2$.$y(-4) = \frac{(-2)^2}{2} - 13 \ln(2) + 13 \ln(2) - 2$.$y(-4) = \frac{4}{2} - 2 = 2 - 2 = 0$.

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

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