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A couple questions on my homework I'm struggling with are as follows. If anyone could help me with these that would be awesome

$$\frac{dy}{dx} + 3y = 12x+19,\qquad y(0)=-4.$$

$$\frac{dy}{dx}=(4x+9y)/3x\qquad y(1)=2$$

Sorry but here's another that I seem to be getting wrong. $$\frac{dx}{dy}=(x−8)e^{−2y} \qquad y(8)=\ln(8)$$

The solution I get is $y=\ln(2x^2-16x+64)/2$ but apparently that's wrong

Thanks

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    Is the third problem **really** $\frac{dx}{dy}$ instead of $\frac{dy}{dx}$?2012-03-24
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    If it is $\frac{dy}{dx}$, then it is separable and you can solve it by simple integration; you get $\int e^{2y}\,dy = \int(x-8)\,dx$, or $e^{2y} = 2(\frac{1}{2}x^2 - 8x + C)$, hence $2y=\ln(x^2 -16x+D)$, which means you messed up your arithmetic again. But if the problem has $\frac{dx}{dy}$, then you did it completely wrong.2012-03-24

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