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

Initial Value Problem

Hint says: Find an integrating factor

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    I still ca$n$$n$ot figure this out.2012-10-04

4 Answers 4

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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.

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    I think with your help, Ryan got the right answer! +12013-03-23
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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)} \,.$

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    So I got the integral, but how would I integrate: e^(coax)y(x)?2012-10-04
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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.

$\dfrac{dy}{dx}=y\sin x-2\sin x$

$\dfrac{dy}{dx}=(y-2)\sin x$

$\dfrac{dy}{y-2}=\sin x~dx$

$\int\dfrac{dy}{y-2}=\int\sin x~dx$

$\ln(y-2)=-\cos x+c$

$y-2=Ce^{-\cos x}$

$y=Ce^{-\cos x}+2$

$y(0)=0$ :

$Ce^{-1}+2=0$

$C=-2e$

$\therefore y=-2ee^{-\cos x}+2=2-2e^{1-\cos x}$

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Hint: Write it as $\frac{y'}{y-2}=\sin(x)$ and integrate.

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    The answer above is correct. However, if the person asking the question insists you use the hint, then you write the ODE as $2012-10-04