MHB Verify My IVP Help Appreciated!

[shamieh]

Need someone to verify that my solution is correct, thanks in advance.

Solve the IVP $y'' - y = e^t$, $y(0) = 0$, $y'(0) = 1$

Solution: $\frac{1}{2}te^t + \frac{1}{2}e^t - \frac{1}{2}e^{-t}$

 

[Ackbach]

Well, your solution solves the DE and the first initial condition, but not the initial condition involving the derivative. I get, for your solution, that $y'(0)=\tfrac32$. Can you show your steps?

 

[shamieh]

$y_h = C_1e^t + C_2e^{-t}$ and I know that $y_p = \frac{1}{2}te^t$

so $y(0) = C_1 + C_2$ and $y'(0) = C_1e^t - C_2e^{-t}$

so
$C_1 + C_2 = 0$
$+$
$C_1 - C_2 = 1$

$C_2 = -1/2$ and $C_1 = 1/2$

so $y_p + y_h = \frac{1}{2} t e^t + \frac{1}{2}e^t - \frac{1}{2}e^{-t}$

 

[shamieh]

So is my solution correct?

 

[MarkFL]

So is my solution correct?

You posted the same solution as in your initial post, which you have already been told is incorrect. :D

You should have found the general solution to the given ODE is:

$$y(t)=c_1e^t+c_2e^{-t}+\frac{t}{2}e^{t}$$

And so we find:

$$y'(t)=c_1e^t-c_2e^{-t}+\frac{1}{2}e^{t}(t+1)$$

Now, to determine the parameters, we use the initial conditions:

$$y(0)=c_1+c_2=0$$

$$y'(0)=c_1-c_2+\frac{1}{2}=1\implies c_1-c_2=\frac{1}{2}$$

Thus, we obtain:

$$\left(c_1,c_2\right)=\left(\frac{1}{4},-\frac{1}{4}\right)$$

Thus, the solution satisfying the given conditions is:

$$y(t)=\frac{1}{4}e^t-\frac{1}{4}e^{-t}+\frac{t}{2}e^{t}=\frac{1}{4e^t}\left((2t+1)e^{2t}-1\right)$$

 

[shamieh]

Oh now I see what you all are saying Mark. I did not put the particular solution with the homogeneous solution & took the derivative incorrectly. Gonna re-work the problem. Thanks so much.

 

[shamieh]

Ahh I got the correct solutuion. Thank you