# Y" -y' -2y = 4(t^2) , y(0) = 1 , y'(0) =4?

1. Mar 14, 2016

### foo9008

1. The problem statement, all variables and given/known data
the answer that i found id 2(e^t) -1 , why it is wrong ? the answer given is cosht

2. Relevant equations

3. The attempt at a solution

2. Mar 14, 2016

### SteamKing

Staff Emeritus
Did you substitute your answer back into the original differential equation? Did it satisfy this equation and all of the initial conditions?

If the answer is no, it didn't, then that's why.

BTW: You should not put one equation in the thread title when the problem covers a completely different equation.

3. Mar 14, 2016

### Staff: Mentor

Furthermore, the equation should be in the problem statement section, not in the thread title.

4. Mar 14, 2016

### LCKurtz

If you try your answer and the book's answer in the original equation, you will see both are incorrect. Either that, or you copied the problem incorrectly. In your work, $\mathcal L(y) \ne s Y(s)$. Also, there is no reason to post this problem as an image instead of just typing it.

5. Mar 16, 2016

### HallsofIvy

Staff Emeritus
Are you required to use the Laplace transform? This is a simple second order, linear equation with constant coefficients. It has characteristic equation $r^2- r- 2= (r- 2)(r+ 1)$ and a "specific solution" to the entire equation must be of the form $Ax^2+ Bx+ C$.

6. Mar 16, 2016

### Staff: Mentor

No. The DE in the title is different from the one in the image that the OP posted. The initial value problem in the posted image is y'' - y = 0, y(0) = 1, y'(0) = 1.

7. Mar 16, 2016

### HallsofIvy

Staff Emeritus
Which is even easier without using the Laplace transform! The characteristic equation is r^2- 1= (r- 1)(r+ 1)= 0.