<p>MrWheezy and I were focusing on a calculus bc problem..</p>
<p>dy/dx = 2y - sin x (2y MINUS sinx!!)</p>
<p>this was introduced to another thread in CC as an FRQ from 1986...
and we are both stuck. I am desperate and I presume so is MrWheezy</p>
<p>I used the integration factor method...</p>
<p>Can anyone help us out?</p>
<p>I sent Mr.Wheezy the solution to part (a), you first need to see that this equation is in GENERAL form.You do have to use the integration factor method. If you want to see a video how to solve this sort of differential equations check this out: [YouTube</a> - Mr Joyce AP Calculus - First Order Linear Differential Equations I](<a href=“http://www.youtube.com/watch?v=kA9xUAF_KOw&feature=related]YouTube”>http://www.youtube.com/watch?v=kA9xUAF_KOw&feature=related)</p>
<p>You should get:
P(x) = -2
Q(x) = -5sin(x)
u(x) = e^(-2x) </p>
<p>The difficult part would be integrating ∫-5[e^(-2x) * sin(x)] dx, you have to do integration by parts. Then multiply(after integration) the expression 1/u(x).</p>
<p>Wait do they even test this on the recent BC exams? Integrating factors?</p>
<p>Btw the solution:
"
Yes, First the we need to recognize that the differential equation is in GENERAL form: dy/dx + P(x)<em>y = Q(x) (our equation: dy/dx = 2y - 5sin(x))
=> dy/dx - 2y = -5sin(x) , Therefore we have P(x)= -2 and Q(x) = -5sin(x).
Now the GENERAL solution</em> for a diffy equation is</p>
<p>y = (1/u(x))<em>∫ [Q(x)</em>u(x)] dx , where u(x) = e^(∫ P(x) dx). </p>
<p>Now to find u(x) first integrate P(x): ∫ -2 dx = -2x (we can leave out the constant of integration here). Now that is your exponent for e: u(x)= e^(-2x).
Plugging in for P(x), Q(x), and u(x) we obtain:</p>
<p>y= (1/e^(-2x))<em>∫ [-5sin(x)</em>e^(-2x)] dx, First integrate (using integration by parts) then multiply by the (1/e^(-2x)) expression. </p>
<p>I hope this was helpful.
"</p>
<p>I don’t think it is on there anymore, but if it is, take a hit of a point, it’s not that big of a deal.</p>
<p>You can use undetermined coefficients or variation of parameters to solve this too.</p>
<p>after struggling with these in my self study time in school i have finally figured this out. Thank you so much lil_killer129 Now having seen this video, I think I know how to do it :)</p>