# $dy/dx + 3y = 8$, $y(0) = 0$ : Initial value problem

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

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$– Stefan Smith Oct 4 '12 at 21:42 This problem will admit separation of variables. – Michael Hardy Oct 4 '12 at 22:16 ## 4 Answers Write this as$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}$- You have$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}}$. - $$\int\frac{y'}{8-3y}dx=\int 1 dx\\ -\frac13\ln(8-3y)=x+c\\ 8-3y=e^{-3(x+c)}\\ y=-\frac13(e^{-3(x+c)}-8),$$ and with$y(0)=0$you'll get$0=-\frac13(e^{-3c}-8)$, which gives$c=-\frac13\ln8$. - 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: en.wikipedia.org/wiki/Integrating_factor –  Marra Oct 4 '12 at 21:47