Differential equation with Laplace transform

Also, when substituting for y(0), it should be y'(0) since the y(0) term is already accounted for in the L[t y''] term.
  • #1
springo
126
0

Homework Statement


[tex]t y'' + (3 t-1) y' + 3 y = 6 e^{-3t}[/tex]
y(0) = 1
y(5) = 2

Homework Equations



The Attempt at a Solution


I tried applying the Laplace transform to the equation but I was having a little trouble...
L [t·y''] = -dL[y'']/ds = s2·Y' + 2·s·Y - y(0) = s2·Y' + 2·s·Y - 1
L [t·y'] = -dL[y']/ds = s·Y' + Y
Therefore:
[tex]s^2 Y' + 2 s Y - 1 + 3 s Y' +3 Y - s Y + 1+ 3 Y = \frac{6}{s+3}[/tex]
[tex](s^2+3s)Y'+(s+6)Y=\frac{6}{s+3}[/tex]

And then solve the Y(s) and then use the inverse transform.
But it doesn't yield the correct result. I did the rest on the computer so I guess it's when I apply the transform where I make a mistake.

Thanks for your help.
 
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  • #2
I think you dropped the negative sign when calculating L[t y''] and L[t y'].
 

What is the Laplace transform?

The Laplace transform is a mathematical tool used to solve differential equations. It converts a function of time into a function of frequency.

Why do we use the Laplace transform to solve differential equations?

The Laplace transform can simplify the process of solving differential equations by converting them into algebraic equations, which are often easier to solve.

How do you use the Laplace transform to solve a differential equation?

To solve a differential equation using the Laplace transform, the differential equation is first transformed into an algebraic equation using the Laplace transform. Then, algebraic methods are used to solve the equation and obtain the solution in terms of the original variable.

What are the advantages of using the Laplace transform to solve differential equations?

The Laplace transform can help to solve more complicated differential equations that may be difficult to solve using traditional methods. It also allows for the use of initial and boundary conditions to be incorporated into the solution.

Are there any limitations to using the Laplace transform to solve differential equations?

The Laplace transform is not always applicable, as some functions may not have a Laplace transform. It also assumes that the differential equation has a unique solution.

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