6.0 KiB
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:
v_{1}'y_{1}+v_{2}'y_{2}=0Compute 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: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 forv_{1}andv_{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 getv_{1}andv_{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:
y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)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}]}Alternatively, you could memorize the system of equations and solve forv_{1}'andv_{2}'. Ie:y_{p}(t)=v_{1}(t)y_{1}(t)+v_{2}(t)y_{2}(t)v_{1}'y_{1}+v_{2}'y_{2}=0v_{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^tWe are done.