Differential Equations Ordinary and Partial Differential Equations Math Forum

 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: 7,671 Thanks: 2651 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. Thanks from Yeep 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: 20,835 Thanks: 2162 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: 7,671
Thanks: 2651

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by Yeep 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?
I would use Variation of Parameters, unless you happen to have a particular solution that works (up to multiplicative constants).

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 non-homogeneous 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: 20,835 Thanks: 2162 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: 20,835 Thanks: 2162 That would correspond to a right-hand side of t³e^(-t), not te^(-t). November 15th, 2016, 02:00 PM   #9
Newbie

Joined: Nov 2016
From: Isenguard

Posts: 6
Thanks: 0

Quote:
 Originally Posted by skipjack That would correspond to a right-hand side of t³e^(-t), not te^(-t).
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: 20,835 Thanks: 2162 What you're looking for doesn't exist if the right-hand side is te^(-t). Tags 2nd, general, ode, order, proving, solution Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post BonaviaFx Trigonometry 1 November 19th, 2015 01:16 PM woo Differential Equations 1 April 27th, 2015 04:51 PM Tooperoo Calculus 4 October 16th, 2013 01:10 AM Linda2 Calculus 3 May 6th, 2012 09:21 PM Infinity Computer Science 3 October 22nd, 2007 05:45 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      