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 October 30th, 2011, 12:08 PM #1 Newbie   Joined: Oct 2011 Posts: 5 Thanks: 0 Variation of parameters? Find the general solution of the differential equation: y'' - 3y' = 5sin(3x) roots are 0 and 3, so how do i get a Wronskian from that? Can't use undetermined coeffs because there's no y term..
 October 30th, 2011, 02:16 PM #2 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Variation of parameters? Don't just memorize formulas (like the "Jacobian"), learn concepts. Here, the fact that 0 and 3 satisfy the characteristic equation means that solutions to the associated homogeneous equation are $e^{3x}$ and $e^{0x}= 1$. From that you can find that the Jacobian is $\left|\begin{array}{cc}e^{3x}=&1 \\ 3e^{3x}=&0\end{array}\right|= -3e^{3x}$. But more important is the idea that we can find a solution to the entire equation of the form $y(x)= u(x)e^{3x}+ v(x)(1)$ for some functions u(x) and v(x). In fact it is easy to see there are an infinite number of such solutions-given a solution, y, pick u(x) to be any differentiable function you please, and you can solve for v(x). If $y= u(x)e^{3x}+ v(x)$, then $y'= u'e^{3x}+ 3ue^{3x}+ v#39;$. Since there exist an infinite number of such solutions, we "narrow the search" by requiring that $u'e^{3x}+ v'= 0$. That leaves $y'= 3ue^{3x}$ so that $y''= 3u#39;e^{3x}+ 9ue^{3x}$. Putting those forms for y' and y'' into the differential equation, we have $y''- 3y'= 3u'e^{3x}+ 9ue^{3x}- 3(3ue^{3x})= 3u#39;e^{3x}= 5sin(3x)$. Solve that for u', then integrate to find u. Once you know u', you can put that into the previous condition, $u'e^{3x}+ v'= 0$ to solve for v' and integrate to find v. However, I don't know why you say "Can't use undetermined coeffs because there's no y term". Whether there is a y term or not is irrelevant. Try y(x)= Asin(3x)+ Bcos(3x).
 October 31st, 2011, 07:01 AM #3 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,989 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Variation of parameters? [color=#000000]Another way to do it. Particular Solution: Substitute $y(x)=c_{1}\sin(3x)+c_2\cos(3x)$ $y''(x)-3y'(x)=5\sin(3x)\Rightarrow -9c_1\sin(3x)-9c_2\cos(3x)+9c_1\cos(3x)-9c_2\sin(3x)=5\sin(3x)\Rightarrow$ $-9\left((c_1+c_2)\cos(3x)+(c_1-c_2)\sin(3x)\right)=5\sin(3x)\Rightarrow \left\{\begin{matrix} c_1+c_2=0 \\ c_{1}-c_{2}=-\frac{5}{9}\end{matrix}\Rightarrow$ $y_{p}=\frac{5}{18}\left(\cos(3x)-\sin(3x)\right)$ Homogeneous Solution: $y''(x)-3y'(x)=0\Rightarrow y'(x)-3y(x)=c_1\Rightarrow \left[e^{-3x}y(x)\right]#39;=c_{1}e^{-3x}\Rightarrow e^{-3x}y(x)=-c_1\frac{e^{-3x}}{3}+c_2\Rightarrow y(x)=-\frac{1}{3}c_1 +c_2 e^{3x}$ so $y(x)=y_{hom}+y_{p}=-\frac{1}{3}c_1 +c_2 e^{3x}+\frac{5}{18}\left(\cos(3x)-\sin(3x)\right)$ .[/color]

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