Finding the Fundamental Solutions of a Third Order ODE

[Temp0]

Homework Statement

Find the fundamental solution to ty''' - y'' = 0

Homework Equations

The Attempt at a Solution

I think I'm missing something really obvious, but I have the characteristic polynomial:
$t\lambda^3 - \lambda^2 = 0$
Solving the equation:
$\lambda^2 (t\lambda - 1) = 0$
I get zero repeated twice, so two of the fundamental solutions are 1 and t.
The third solution is $\frac {1}{t}$ to solve the equation, but substituting it in, I get the fundamental solution as e.
However, the right solution is $t^3$.
Did I miss something? Thank you in advance.

 

[ehild]

It is not a constant-coefficient linear de, with solutions in form $e^{\lambda t}$, with $\lambda$ a constant. Lambda can not be 1/t.

You can replace y"=u, a new function and solve for u(t). Then integral twice.

ehild

 

[SteamKing]

Homework Statement

Find the fundamental solution to ty''' - y'' = 0

Homework Equations

The Attempt at a Solution

I think I'm missing something really obvious, but I have the characteristic polynomial:
$t\lambda^3 - \lambda^2 = 0$
Solving the equation:
$\lambda^2 (t\lambda - 1) = 0$
I get zero repeated twice, so two of the fundamental solutions are 1 and t.
The third solution is $\frac {1}{t}$ to solve the equation, but substituting it in, I get the fundamental solution as e.
However, the right solution is $t^3$.
Did I miss something? Thank you in advance.

Yes. The characteristic polynomial method only works for ODEs with constant coefficients. t is not a constant here.

 

[Temp0]

Thanks for all your help! So I ended up solving the first order linear, and getting u(t) = t, which means y'' = t. Taking the integral twice, I'm getting:

y = $\frac {t^3}{6} + C_1 t + C_2$

I kind of understand what's happening now, so $\frac {1}{6}$ is my $C_3$?

 

[LCKurtz]

Thanks for all your help! So I ended up solving the first order linear, and getting u(t) = t, which means y'' = t. Taking the integral twice, I'm getting:

y = $\frac {t^3}{6} + C_1 t + C_2$

I kind of understand what's happening now, so $\frac {1}{6}$ is my $C_3$?

You didn't show your work so It's hard to tell, but I think you are missing a constant in your work to get $u(t)=t$.

There is another way to solve the problem that you might find interesting. Multiply your equation through by $t^2$ then look for a solution in the form $y=t^n$ for unknown $n$ and see what $n$'s work.

 

[willem2]

y = t36+C1t+C2 \frac {t^3}{6} + C_1 t + C_2

I kind of understand what's happening now, so 16 \frac {1}{6} is my C3 C_3 ?

No. The original equation is a linear equation, so if f(t) is a solution, so is Cf(t). Your $C_3$ is just $C_3$.

 

[LCKurtz]

No. The original equation is a linear equation, so if f(t) is a solution, so is Cf(t). Your $C_3$ is just $C_3$.

If you mean what I think you mean, you are wrong. His $t^3/6$ should be $C_3t^3$.

 

[ehild]

Thanks for all your help! So I ended up solving the first order linear, and getting u(t) = t,
which means y'' = t.

The integration constant is missing here. U(t)=C t.

Taking the integral twice, I'm getting:

$y = \frac {t^3}{6} + C_1 t + C_2$

It would be $y = C\frac {t^3}{6} + C_1 t + C_2$

I kind of understand what's happening now, so $\frac {1}{6}$ is my $C_3$?

You can take C₃=C/6, and then the general solution is y=C₁t+C₂+C₃t³.
Each term of the sum is a solution.