I know that one can use an ansatz to determine that $te^{rt}$ is a solution to the ODE $a\ddot{x}+b\dot{x}+cx=0$ where $r$ is the double root of the characteristic polynomial. Is there a way this can be derived explicitly without such an "educated guess"?
Asked
Active
Viewed 138 times
0
-
@Moo I meant $cx$ (with $d=0$). I'll fix that now. – Andrew Paul Sep 19 '19 at 21:34
-
Variation of parameters. – Adrian Keister Sep 19 '19 at 21:35
2 Answers
2
If we let the general solution have the form $ x(t) = y(t) e^{rt} $, then $$ \begin{align} 0 &= a\ddot{x} + b\ddot{x} + cx \\ &= ( a(\ddot{y} + 2r\dot{y} + r^2 y ) + b(\dot{y}+ry) + cy )e^{rt} \\ 0&= a\ddot{y} + (2ra+b)\dot{y} + (ar^2+br+c)y \end{align} $$ If $r$ is a simple root of $ar^2+br+c$, the equation reduces to $$ a\ddot{y} + (2ra+b)\dot{y} = 0 , $$ which we can integrate once to get $$ a\dot{y} + (2ra+b)y = A $$ for some constant $A$. Now, if $A$ is a double root, we also have $2ra+b = 0$, either from multiplying out the brackets in $a(x-r)^2$ and equating coefficients, or using that the derivative of the quadratic also vanishes at a repeated root. Either way, the equation now reduces to $$ a\dot{y} = A , $$ which integrates to $ay=At+B$, which on resubstituting gives the general solution for $x$. A similar approach works for any linear differential equation where you know one solution: this process is called reduction of order.
Lutz Lehmann
- 126,666
Chappers
- 67,606
2
The strategy is to eliminate the exponential factor, which can be done as follows:
The equation with a double root can be brought into the form $$x''-2rx'+r^2x=0$$ Now set $y=x'-rx$ so that $$y'-ry=0$$ whence $y=Ae^{rt}$
We are then faced with $$x'-rx=Ae^{rt}$$ or $$x'e^{-rt}-rxe^{-rt}=A$$
Now set $z=xe^{-rt}$ so that $z'=-rz+x'e^{-rt}$ and we have $$z'+rz-rz=z'=A$$
And the solution to $z'=A$ is $z=At+B$ whence $x=Ate^{rt}+Be^{rt}$
Mark Bennet
- 100,194