Question

Find the solution of the following initial value problem. y'' (t) = 6te! y(0) = 3, y'(0) = 1 y(t) =

Answer 1

1 solution

and y(0) =3 and y' (0)=1

y" = 6te e

taking laplace transform both side

$s^2y(s)-sy(0)-y'(0)=6(\frac{1}{(s-1)^2})$

$s^2y(s)-3s-1=\frac{6}{(s-1)^2}$

$s^2y(s)=\frac{6}{(s-1)^2}+3s+1$

$y(s)=\frac{6}{s^2(s-1)^2}+\frac{3s}{s^2}+\frac{1}{s^2}$

$y(s)=\frac{6}{s^2(s-1)^2}+\frac{3}{s}+\frac{1}{s^2}$

also $\frac{6}{s^2(s-1)^2}=\frac{12}{s}+\frac{6}{s^2}-\frac{12}{(s-1)}+\frac{6}{(s-1)^2}$

$\therefore y(s)=\frac{15}{s}+\frac{7}{s^2}-\frac{12}{(s-1)}+\frac{6}{(s-1)^2}$

Now on taking inverse laplace transform both side

$L^{-1} y(s)=L^{-1}(\frac{15}{s}+\frac{7}{s^2}-\frac{12}{(s-1)}+\frac{6}{(s-1)^2})$

$\therefore y(t)=15+7t-12e^t+6te^t$

Note : we are instructed to do one question at one time