# $dy/dx = y \sin x-2\sin x$, $y(0) = 0$ — Initial Value Problem

$$\frac{dy}{dx} = y\sin x-2\sin x, \quad y(0) = 0.$$

Initial Value Problem

Hint says: Find an integrating factor

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Welcome to this Math Stack Exchange. Notice that your "question" is kind of a rude implication that the community is just expected to compute your answer, like a robot. It's more helpful if you actually ask a question, and give as much information as you can about why you are stuck. For example, do you not know what an "integrating factor " is? –  alex.jordan Oct 4 '12 at 21:05
No I do not. What is an integrating factor? –  Ryan Oct 4 '12 at 21:06
I still cannot figure this out. –  Ryan Oct 4 '12 at 21:30

Hint: Write it as $\frac{y'}{y-2}=\sin(x)$ and integrate.
The answer above is correct. However, if the person asking the question insists you use the hint, then you write the ODE as $– Stefan Smith Oct 4 '12 at 21:06 When we have a linear differential equation$y'+P(x)y=Q(x)$of order 1 then$\mu (x)=e^{\int P(x) dx}$will be an integrating factor. In your case, it is$\mu(x)=e^{\cos(x)}$. Now multiply it both sides of the equation and you see $$d(e^{\cos(x)}y)=-2\sin(x)e^{\cos(x)}$$ The rest is easy. - What do I do next? – Ryan Oct 4 '12 at 21:15 @Ryan: Did you get the right answer? :-) – Babak S. Oct 5 '12 at 7:17 I think with your help, Ryan got the right answer! +1 – amWhy Mar 23 '13 at 0:44 If a differential equation has the form $$y'(x)+p(x)y(x)=q(x)\,, \quad (1),$$ then the integrating factor is given by $$m(x)= {\rm e}^{\int p(x) dx}\,.$$ You need to write your ode in the form of$(1)$and then find the integrating factor. Then just multiply the original equation by the interating factor and integrate $$(m(x)y)'= q(x) \Rightarrow \frac{d}{dx}(e^{\cos(x)}y)=-2\sin(x)e^{\cos(x)}$$ $$\Rightarrow e^{\cos(x)}y(x)=2\int e^{\cos(x)}(-\sin(x))dx + C = 2e^{\cos(x)} +C$$ $$y(x)= 2 + C \,{\rm e}^{-\cos(x)} \,.$$ To find$C$, you need to use the initial condition$y(0)=0$, $$y(0) = 0 = 2 + C{\rm e}^{-\cos(0)} \Rightarrow C = -2 {\rm e}$$ Substituting the value of$C$in the solution gives $$y(x)= 2 - 2 \,{\rm e}^{1-\cos(x)} \,.$$ - So I got the integral, but how would I integrate: e^(coax)y(x)? – Ryan Oct 4 '12 at 21:20 The simplest type of this ODE is that separable rather than linear. So it is not necessary to treat it as a linear ODE to solve.$\dfrac{dy}{dx}=y\sin x-2\sin x\dfrac{dy}{dx}=(y-2)\sin x\dfrac{dy}{y-2}=\sin x~dx\int\dfrac{dy}{y-2}=\int\sin x~dx\ln(y-2)=-\cos x+cy-2=Ce^{-\cos x}y=Ce^{-\cos x}+2y(0)=0$:$Ce^{-1}+2=0C=-2e\therefore y=-2ee^{-\cos x}+2=2-2e^{1-\cos x}\$