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$$\begin{cases} \frac{dy}{dx} + 3y = 8, \\ y(0) = 0. \end{cases}$$

So, I have been getting an answer of $3$ by integrating and getting $\ln(8-3y) = x$ and solving. But my book says the answer must be expressed as a function of $x$. I do not know what to do.

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The standard procedure is: you find a particular solution to the ODE (in this case there is a constant solution), and the general solution to the corresponding homogeneous ODE, which here has the form $y = c*something(x)$. Then you find $c$ so c*something(0) + [particular solution](0) = 0$ – Stefan Smith Oct 4 '12 at 21:42
This problem will admit separation of variables. – Michael Hardy Oct 4 '12 at 22:16

You have $y'(x)+3y(x)=8\Rightarrow \mathbb{e}^{3x}y'(x)+3\mathbb{e}^{3x}y(x)=8\mathbb{e}^{3x}\Rightarrow \mathbb{e}^{3x}y'(x)+\left(\mathbb{e}^{3x}\right)'y(x)=8\mathbb{e}^{3x}\Rightarrow$

$\left(\mathbb{e}^{3x}y(x)\right)'=8\mathbb{e}^{3x}\Rightarrow {\Large\int}\left(\mathbb{e}^{3x}y(x)\right)'\;\mathbb{d}x={\Large\int}8\mathbb{e}^{3x}\;\mathbb{d}x\Rightarrow \mathbb{e}^{3x}y(x)=\frac{8}{3}\left(\mathbb{e}^{3x}\right)+c\Rightarrow $ *for $c\in\mathbb{R}$

$y(x)=\frac{8}{3}+\frac{c}{\mathbb{e}^{3x}}$, now since $y(0)=0\Rightarrow c=-\frac{8}{3}$ and $y(x)=\frac{8}{3}-\frac{8}{3\mathbb{e}^{3x}}$.

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Multiply both sides of the differential equation by $e^{3x}$. This will give you $e^{3x}\dfrac{dy}{dx}+3 e^{3x}y=8e^{x}$, which is $$ \dfrac{d}{dx}(ye^{3x})=8e^{3x} $$ Integrate both sides with respect to x (don't forget to add a constant).

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The method I described is a very common strategy to deal with this kind of problem. You can read more about it here: – Marra Oct 4 '12 at 21:47

$$ \int\frac{y'}{8-3y}dx=\int 1 dx\\ -\frac13\ln(8-3y)=x+c\\ 8-3y=e^{-3(x+c)}\\ y=-\frac13(e^{-3(x+c)}-8), $$ and with $y(0)=0$ you'll get $0=-\frac13(e^{-3c}-8)$, which gives $c=-\frac13\ln8$.

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Write this as $dy/dx+3(y-8/3)$ and then integrate $dy/(y-8/3)+3dx=0$ to get $\ln(y-8/3)+3x=c$ so $y-3/8=Ke^{-3x}$; $y(0)=0$ so $K=-3/8$ and $y=3/8-3/8e^{-3x}$

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