November 14th, 2016, 07:26 PM  #1 
Newbie Joined: Nov 2016 From: Isenguard Posts: 6 Thanks: 0  Proving general solution for 2nd order ODE
Show that y(t) = A(t^2) + Bt, where A and B are arbitrary constants, is the general solution of the differential equation (t^2)y′′ − 2t y′ + 2 y = 0. I got the roots to be r= (1/t)+(i/t) .... Which I would then plug into the equation involving complex roots which is e^(gt)(Acos(ut)+Bsin(ut)) When I solve for g and u and I simplify I do not end up with the general soluton being At^2+Bt.... Help Last edited by Yeep; November 14th, 2016 at 07:37 PM. 
November 14th, 2016, 07:37 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra 
You don't use that characteristic equation for an Euler equation. There are two easier approaches for this, and one harder. The easiest is to determine $y'$ and $y''$ from the given general solution, $y$, and then show that such $y$, $y'$ and $y''$ satisfy the given equation. Alternatively, you can trial the solution $y=x^r$ to get a quadratic for $r$. Finally, you can write $x = \ln t$ (I think), complete the substitution and solve the resulting equation with constant coefficients. Last edited by v8archie; November 14th, 2016 at 07:41 PM. 
November 14th, 2016, 09:26 PM  #3 
Newbie Joined: Nov 2016 From: Isenguard Posts: 6 Thanks: 0 
Taking the same ODE but set it = to te^(t)... with IVP y(0)=y'(0)=0. Would I use the given y(t)+At^2+Bt as my general solution, or do I need to use the method of undetermined Coeff.? Can anyone nudge me in the right direction? 
November 14th, 2016, 11:38 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 17,707 Thanks: 1356 
For t ≠ 0, dividing t²y'' − 2ty' + 2y = 0 by t³ gives (1/t)y'' − (2/t²)y' + (2/t³)y = 0. Integrating that gives (1/t)y'  (1/t²)y = A, where A is a constant. Integrating that gives (1/t)y = At + B, where B is a constant. Multiplying by t gives y = At² + Bt. Allowing t = 0 leads to y = At² + Bt as the general solution, because the existence of y'' at t = 0 implies A and B cannot change at t = 0. 
November 15th, 2016, 02:39 AM  #5  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra  Quote:
I'm sure your guess is incorrect for the particular solution because it is the general solution of the homogeneous equation. Your general solution of the nonhomogeneous equation will be $y(t)=y_p(t) + At^2 + Bt$, but you need to find a suitable $t$. If you chose the third method to solve the homogeneous equation, you could use that transformation to find a particular solution in $x$ by the Method of Undetermined Coefficients.  
November 15th, 2016, 05:34 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 17,707 Thanks: 1356 
There isn't a particular solution without recourse to a special function or its equivalent.

November 15th, 2016, 11:09 AM  #7 
Newbie Joined: Nov 2016 From: Isenguard Posts: 6 Thanks: 0 
Using Variation of Parameters I get that Yp=te^(t). So now all I would need to do is add that to Yh ( which is the general solution) since y=Yp+Yh and solve using my IVP to find the constants A and B?

November 15th, 2016, 01:33 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 17,707 Thanks: 1356 
That would correspond to a righthand side of t³e^(t), not te^(t).

November 15th, 2016, 02:00 PM  #9 
Newbie Joined: Nov 2016 From: Isenguard Posts: 6 Thanks: 0  I'm assuming I can't just look at the general solution and take y1=t^2 and y2=t. Then finding the wronks then using Yp=U1Y1+U2Y2? (where u=W1,2/W)
Last edited by skipjack; November 15th, 2016 at 02:46 PM. 
November 15th, 2016, 02:48 PM  #10 
Global Moderator Joined: Dec 2006 Posts: 17,707 Thanks: 1356 
What you're looking for doesn't exist if the righthand side is te^(t).


Tags 
2nd, general, ode, order, proving, solution 
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