2nd Order Differential Equation (Complex)

[roam]

Homework Statement

Find the general solution to the following differential equation:

$\frac{d^2 y}{dt^2} - 2 \frac{dy}{dt} + 2y =e^t$

The correct answer must be: $y(t) = C_1 e^t \cos t + C_2 e^2 \sin t +e^t$

The Attempt at a Solution

I haven't been able to get the correct answer so far. The eigenvalues are:

$\lambda^2-2 \lambda +2 = 0$

$\lambda = \frac{2 \pm \sqrt{4-8}}{2} = 1 \pm i$

So the general solution of the corresponding homogeneous equation is:

$y_h(t) = C_1 e^{(1-i)t} + C_2 e^{(1-i)t}$

And with inspection a particular solution is y_p = e^t, so the general solution must be:

$y(t) = C_1 e^{(1-i)t} + C_2 e^{(1-i)t} + e^t$

Using Euler's formula:

$y(t) = C_1 \frac{e^t}{\cos t + i \sin t} + C_2 e^t (\cos t + i \sin t)+ e^t$

So, why is my answer so different from the correct answer provided? What do I have to do in order to get the correct answer?

Any help is really appreciated.

 

[DryRun]

The general solution of the reduced equation should be:
$$y_c=e^{t}(C_1 \cos t + C_2 \sin t)$$Then, use the method of variation of parameters to find the particular integral.

The correct answer must be: $y(t) = C_1 e^t \cos t + C_2 e^2 \sin t +e^t$

I think you mistyped the answer as i got the general equation: $y(t) = C_1 e^t \cos t + C_2 e^t \sin t +e^t$

 

[LCKurtz]

Homework Statement

Find the general solution to the following differential equation:

$\frac{d^2 y}{dt^2} - 2 \frac{dy}{dt} + 2y =e^t$

The correct answer must be: $y(t) = C_1 e^t \cos t + C_2 e^2 \sin t +e^t$



$\lambda = \frac{2 \pm \sqrt{4-8}}{2} = 1 \pm i$

So the general solution of the corresponding homogeneous equation is:

$y_h(t) = C_1 e^{(1-i)t} + C_2 e^{(1-i)t}$

And with inspection a particular solution is y_p = e^t, so the general solution must be:

$y(t) = C_1 e^{(1-i)t} + C_2 e^{(1-i)t} + e^t$

Using Euler's formula:

$y(t) = C_1 \frac{e^t}{\cos t + i \sin t} + C_2 e^t (\cos t + i \sin t)+ e^t$

The homogeneous solution should be written$$y_h(t) = C_1 e^{(1+i)t} + C_2 e^{(1-i)t} + e^t
=C_1e^te^{it}+C_2e^te^{-it} =
C_1e^t(\cos t + i \sin t)+C_2e^t(\cos t - i\sin t)$$from which you can get the sine - cosine form by collecting terms and renaming the constants.

 

[roam]

Yest, that was a typo I meant e^t not e².

Thanks a lot LCKurtz for the hint, I finally got the correct expression.