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Sagot :
The Laplace of the given equation is [tex]y=-\frac{15e^2}{8}e^{2t}+\left(-3e^2+\frac{15e^2}{4}\right)e^{2t}t+\frac{e^2t^4}{4}+e^2t^3+\frac{9e^2t^2}{4}+3e^2t+\frac{15e^2}{8}[/tex].
According to the statement
we have given that the equation and we have to solve this problem with the help of the Laplace transform.
So, According to the statement
the given equation is
y'' − 4y' + 4y = t^3e^2t, y(0) = 0, y'(0) = 0
And the Laplace transform is an integral transform method which is particularly useful in solving linear ordinary differential equations.
Firstly fill the value of the Laplace of the second order derivative and put x = 0 in the equation
And same it with the first order of the derivative.
And then the Laplace of the equation become
[tex]y=-\frac{15e^2}{8}e^{2t}+\left(-3e^2+\frac{15e^2}{4}\right)e^{2t}t+\frac{e^2t^4}{4}+e^2t^3+\frac{9e^2t^2}{4}+3e^2t+\frac{15e^2}{8}[/tex]
So, The Laplace of the given equation is [tex]y=-\frac{15e^2}{8}e^{2t}+\left(-3e^2+\frac{15e^2}{4}\right)e^{2t}t+\frac{e^2t^4}{4}+e^2t^3+\frac{9e^2t^2}{4}+3e^2t+\frac{15e^2}{8}[/tex]
Learn more about Laplace transform here
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