Solve $$y''' - 3y'' + 3y' -y = t^2e^t $$
when $$y''(0) = -2\,\quad, \quad y'(0) = 0\quad, \quad y(0) = 1$$
EDIT: I arrived at $$y = 2/(s-1)^6 - 1/(s-1)^3$$
EDIT2: Turns out I forgot to multiply the s terms
Laplace differential equation with initial conditions
0
$\begingroup$
ordinary-differential-equations
laplace-transform
initial-value-problems
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0solve in the first step the equation $$y'''-3y''+3y'-y=0$$ – 2017-01-20
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1make the Ansatz $$y=e^{\lambda t}$$ – 2017-01-20
1 Answers
1
Hint
Apply
Laplace Transform of the Derivative $$\mathcal{L}[y^{(n)}](s)=s^nY(s)-s^{n-1}y(0)-s^{n-2}y'(0)-\cdots s\,y^{(n-2)}(0)-y^{(n-1)}(0)$$
Derivatives of the Laplace Transform
$$\mathcal{L}[{t^ny(t)}](s)=(-1)^n\frac{d^n}{ds^n}(Y(s))$$
Table of Laplace Transforms $$\mathcal{L}[{e^t}](s)=\frac{1}{s-1}$$