I am trying to solve a third order differential equation problem with laplace transform. But I am stuck since 3 days... Could someone tell me what I did incorrectly?
I transformed my equation in the s-domain. But according to all online calculators (ex: wolframalpha.com), there is no inverse laplace possible....
the equation(+ first steps): https://www.dropbox.com/s/b4dt8cqzkjk996w/IMAG1178.jpg?dl=0
$$y'''(t)-3y'(t)+2y(t)=te^{-2t}\Longleftrightarrow$$
The general solution willbe the sum of the complementary solution and particular solution:
$$\frac{\text{d}^3y(t)}{\text{d}t^3}-3\frac{\text{d}y(t)}{\text{d}t}+2y(t)=0\Longleftrightarrow$$
Assume a solution will be proportional to $e^{\lambda t}$ for some constant $\lambda$. Substitute $y=e^{\lambda t}$:
$$\frac{\text{d}^3}{\text{d}t^3}\left(e^{\lambda t}\right)-3\frac{\text{d}}{\text{d}t}\left(e^{\lambda t}\right)+2e^{\lambda t}=0\Longleftrightarrow$$
Substitute $\frac{\text{d}^3}{\text{d}t^3}\left(e^{\lambda t}\right)=\lambda^3e^{\lambda t}$ and $\frac{\text{d}}{\text{d}t}\left(e^{\lambda t}\right)=\lambda e^{\lambda t}$:
$$\lambda^3e^{\lambda t}-3\lambda e^{\lambda t}+2e^{\lambda t}=0\Longleftrightarrow$$ $$\left(\lambda^3-3\lambda+2\right)e^{\lambda t}=0\Longleftrightarrow$$
Since $e^{\lambda t}\ne 0$ for finite $\lambda$, the zros must come from the polynomial $\lambda^3-3\lambda+2=0$:
$$\lambda^3-3\lambda+2=0\Longleftrightarrow$$ $$\left(\lambda-1\right)^2\left(\lambda+2\right)=0\Longleftrightarrow$$ $$\lambda=-2\space\space\vee\space\space\lambda=1\space\space\vee\space\space\lambda=1\Longleftrightarrow$$
The root $\lambda=-2$ gives $y_1(t)=c_1e^{-2t}$ as a solution, where $c_1$ is an arbitrary constant. The multiplicity of the root $\lambda=1$ is $2$ which gives $y_2(t)=c_2e^{t}$ and $y_3(t)=c_3e^{t}t$ as solutions, where $c_2$ and $c_3$ are arbitrary constants. The general solution is the sum of the above solutions:
$y(t)=y_1(t)+y_2(t)+y_3(t)=c_1e^{-2t}+c_2e^{t}+c_3e^{t}t$
$$y(t)=\frac{e^{-2t}\left(t(4+3t)+54c_1\right)}{54}+e^t\left(c_2+tc_3\right)$$