Higher Order Differential Equations, Solutions related

[NCyellow]

Homework Statement

Given the following differential equation
t^3y''' - t^2y'' + 2ty' - 2y = 0; t > 0
Find a solution that satisfies differential equation and the initial conditions
y(1) = 3; y'(1) = 2; y''(1) = 1

Homework Equations

The Attempt at a Solution

I tried plugging in 1 for y, y' and y'', but that gave me t^3y'''-t^2+2t-2=0, which doesn't give me anything... i looked in the book but still am positively stumped... Please help.

 

[HallsofIvy]

I don't understand why you would expect that to work! That would give you the value of y"' at x= 1 but no where else. The fact that they are equal to 1 at x= 1 tells you nothing about their values in general.

That equation simply says that
$$\frac{d^3y}{dt^3}= \frac{t^2- 2t+ 2}{t^3}= t^{-1}- 2t^{-2}+ 2t^{-3}$$

Integrate three times.

 

[NCyellow]

I'm sorry, how did you get that equation?

 

[elibj123]

Are you familiar with constant coefficient ode's?

If yes, then make the substitution x=ln(t), and find u(x)=y(e^x).
(Remember that the derivatives of y wrt to t are different, and use the chain rule)

 

[LCKurtz]

This type of DE is called an Euler or Euler-LaGrange equation. Look for a solution of the form y = tⁿ. When you plug that into the DE you will get an equation that you can solve for n. It is similar to what happens for constant coefficient DE's. You can read about them, for example at:

http://tutorial.math.lamar.edu/Classes/DE/EulerEquations.aspx

In particular, that location shows you what happens if you get a repeated root for n.

 

[LCKurtz]

That equation simply says that
$$\frac{d^3y}{dt^3}= \frac{t^2- 2t+ 2}{t^3}= t^{-1}- 2t^{-2}+ 2t^{-3}$$

Integrate three times.

I'm sorry, how did you get that equation?

Sometimes we all type before we look closely. I don't think Halls really meant that.