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Variation of parameters

Professors definition/derivations during lecture <- I found this to be too big brain for me. Here is a simplified definition:

Variation of parameters is a method to solve:

$ay''+by'+cy=f(t)$

#voparam Variation of parameters is generally known to be a more powerful replacement for method of undetermined coefficients. First, find the homogenous solution: y_{h}=c_{1}y_{1}(t)+c_{2}y_{2}(t) Now we need the particular solution, let y_{p} be in the following form: y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t) <- btw y_{1} and y_{2} are often called a fundamental pair. They are obtained from your homogenous solution. Impose the following:

  1. v_{1}'y_{1}+v_{2}'y_{2}=0 Compute the derivatives and simplify: y_{p}'=v_{1}y_{1}'+v_{2}y_{2}' y_{p}''=v_{1}'y_{1}'+v_{1}y_{1}''+v_{2}'y_{2}'+v_{2}y_{2}'' Now we plug those into the second order equation and simplify:
  2. v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a} We now have a system of two equations (1 and 2). Now we can solve for v_{1} and v_{2}: Using Cramer's rule, we can solve for the system of equations and obtain the solutions: v_{1}'=-\frac{{f(t)y_{2}(t)}}{aW[y_{1},y_{2}]}; v_{2}'=\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]} <- integrate both sizes to get v_{1} and v_{2}. When integrating, you don't need to add a generic constant. also, W[y_1,y_{2}] is the Wrońskian, and it equals to: \det \begin{pmatrix}y_{1}&y_{2}\\ y_{1}' &y_{2}'\end{pmatrix}=y_{1}y_{2}'-y_{2}y_{1}' Finally, the general solution is:
y(t)=y_{h}+y_{p}\qquad \text{where}\qquad y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)

What you need to remember

#remember So, what do you need to commit to memory? I believe memorizing these three is a good tradeoff between memory allocated and speed for when you're solving a #voparam problem:

  1. y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)
  2. v_{1}'=-\frac{{f(t)y_{2}(t)}}{aW[y_{1},y_{2}]}
  3. v_{2}'=\frac{{f(t)y_{1}(t)}}{aW[y_{1},y_{2}]} Alternatively, you could memorize the system of equations and solve for v_{1}' and v_{2}'. Ie:
  4. y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)
  5. v_{1}'y_{1}+v_{2}'y_{2}=0
  6. v_{1}'y_{1}'+v_{2}'y_{2}'=\frac{f(t)}{a} This is what the prof likes. I love you Dr. Minev, but this I personally disagree with. I think I'll stick with the formulas.

#ex #second_order #IVP #voparam #mouc Solve the IVP:

y''+4y=2\tan(2t)-e^t \qquad y(0)=0 \qquad y'(0)=\frac{4}{5}

Can we use undetermined coefficients? Yes and no. We can use it on the e^t term. However, guessing and checking y_{p} for the 2\tan(2t) term might take a really, really long time. First, find general solution to homogenous counterpart: y''+4y=0 -> r^2+4=0 -> r_{1,2}=\pm 2i y_{h}(t)=c_{1}\cos(2t)+c_{2}\sin(2t) Done. Easy peasy. For -e^t lets use method of undetermined coefficients: y''+4y=-e^t y_{p_{1}}(t)=Ae^{t} 5Ae^t=-e^t A=-\frac{1}{5} y_{p_{1}}(t)=-\frac{1}{5}e^t

Now for 2\tan(2t), we cannot realistically use method of undetermined coefficients. Let's use variation of parameters instead: y''+4y=2\tan(2t) y_{p_{2}}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t) where y_{1}=\cos(2t), y_{2}=\sin(2t) recall: v_{1}'=-\frac{f(t)y_{2}}{aW[y_{1},y_{2}]} v_{2}'=\frac{f(t)y_{1}}{aW[y_{1},y_{2}]} plugging in: v_{1}'=-\frac{2\tan(2t)\sin(2t)}{(1)(\cos2t(2)\cos2t-\sin2t(-2)\sin2t}=-\frac{2\tan(2t)\sin(2t)}{2\cos^22t+2\sin^22t}=-\tan(2t)\sin(2t) v_{2}'=\frac{2\tan(2t)\cos(2t)}{2}=\sin(2t) <- isn't it nice how we can reuse our computation for the denominator? :D Now we integrate. v_{2}=-\frac{1}{2}\cos(2t) <- Don't add a constant of integration, we want one solution only. v_{1}=-\int \frac{\sin^2(2t)}{\cos(2t)} \, dt v_{1}=-\int \frac{{1-\cos^2(2t)}}{\cos(2t)} \, dt v_1=-\int sec(2t) \, dt+\int \cos(2t) \, dt v_{1}(t)=-\frac{1}{2}\ln\mid sec(2t)+\tan(2t)\mid+\frac{1}{2}\sin(2t) y_{p_{2}}(t)=v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)

^Now you can start to see how guessing y_{p_{2}} would take a really, really long time.

y(t)=y_{h}(t)+y_{p_{1}}(t)+y_{p_{2}}(t) =c_{1}\cos(2t)+c_{2}\sin(2t)+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)-\frac{1}{5}e^t is the general answer.

IVP solution: y(0)=0=c_{1}-y_{p}(0)=c_{1}-\frac{1}{5} \implies c_{1}=\frac{1}{5} y'(0)=\frac{4}{5}=2c_{2}+v_{1}'(0)+2v_{2}(0)-\frac{1}{5} \implies c_{2}=1

y(t)=\frac{1}{5}\cos(2t)+\sin(2t)-\frac{1}{5}e^t+v_{1}(t)\cos(2t)+v_{2}(t)\sin(2t)

#end of lecture 9 #start of lecture 10 #ex #second_order #voparam #mouc Find the general solution for:

y''-2y'+y=e^t\ln(t)+2\cos(t)

Find homogenous solution first: r^2-2r+1=0 r_{1,2}=1 (repeated root) y_{h}(t)=c_{1}e^t+c_{2}te^t 2) y_{p}(t)=? y''-2y'+y=2\cos (t) let's use method of undetermined coefficients: y_{p_{1}}=A\cos(t)+B\sin(t) is our guess y_{p_{1}}'=-A\sin t+B\cos t y_{p_{1}}''=-A\cos t-B\sin t -A\cos t-B\sin t+2A\sin t-2B\cos t+A\cos t+B\sin t=2\cos t -2B\cos t+2A\sin(t)=2\cos t(t) \implies A=0,\ B=-1 y_{p_{1}}=-\sin(t) y''-2y'+y=e^t\ln(t) cant use undetermined coefficients, use variation of parameters y_{p}''(t)=v_{1}y_{1}+v_{2}y_{2} =v_{1}e^t+v_{2}te^t Compute v_{1} and v_{2}. This time let's do it using the linear system for practice: eq1) e^tv_{1}'+te^tv_{2}'=0 eq2) e^tv_{1}'+(te^t+e^t){v_{2}'}=e^t{\ln t} subtract eq1 from eq2: v_{2}'=\ln(t) v_{2}(t)=\int \ln(t) \, dt integrate by parts =t\ln(t)-\int t\frac{1}{t} \, dt =t\ln(t)-t no constant of integration. compute v_{1} now: v_{1}'=-tv_{2}' =-t\ln t integrate to get v_1: v_{1}=-\int t\ln t \, dt integrate by parts (btw integration by parts will be the most important integration technique in this course): v_{1}=-\frac{1}{2}(t^2\ln t)-\int t^2\frac{1}{t} \, dt =-\frac{1}{2}\left( t^2\ln t-\frac{t^2}{2} \right)=-\frac{1}{2}t^2\ln t+\frac{1}{4}t^2 y_{p}''(t)=(\frac{1}{2}t^2\ln t+\frac{1}{4}t^2)e^t+(t\ln t-t)te^t y_{p}(t)=-\sin(t)+\frac{1}{2}t^2\ln(t)e^t-\frac{3}{4}t^2e^t general solution is produced by adding the homogenous eq with y_{p}(t) general solution:

y(t)=c_{1}e^t+c_{2}te^t-\sin(t)+\frac{1}{2}t^2\ln(t)e^t-\frac{3}{4}t^2e^t

We are done.