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 November 14th, 2016, 08:26 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 08:37 PM.
 November 14th, 2016, 08:37 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,598 Thanks: 2583 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 08:41 PM.
 November 14th, 2016, 10:26 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 15th, 2016, 12:38 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,263 Thanks: 1958 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, 03:39 AM   #5
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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, 06:34 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,263 Thanks: 1958 There isn't a particular solution without recourse to a special function or its equivalent.
 November 15th, 2016, 12:09 PM #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, 02:33 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,263 Thanks: 1958 That would correspond to a right-hand side of t³e^(-t), not te^(-t).
November 15th, 2016, 03:00 PM   #9
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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 03:46 PM.

 November 15th, 2016, 03:48 PM #10 Global Moderator   Joined: Dec 2006 Posts: 20,263 Thanks: 1958 What you're looking for doesn't exist if the right-hand side is te^(-t).

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

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