Solve the following differential equation by the Laplace transform method:
$y''+y'= 8x^2, \\ y(0)=0 , \\ y'(0)=1$
By applying the Laplace transform to both sides:
$L\{y''+y'\}=L\{8x^2\}$
$L\{y''+y'\}=s^2L\{y\}-sy(0)-y'(0)+sL\{y\}-y'(0)$
$L\{8x^2\}=\frac{16}{s^3}$
Then
$s^2L\{y\}-sy(0)-y'(0)+sL\{y\}-y(0)= \frac{16}{s^3}$
When entering initial conditions $y(0)=0 \\ y'(0)=1$
$s^2L\{y\}- s(0)-1+sL\{y\}-0=\frac{16}{s^3}$ By simplifying
$s^2L\{y\}+sL\{y\} -1 =\frac{16}{s^3}$
When solving for $L\{y\}: L\{y\}= \frac{16+s^3}{s^4(1+s)}$
Applying the inverse Lapalace transform:
$y= L^{-1} \{ \frac{16+s^3}{s^4(1+s)}\}$
$y= -15+16x-8x^2+ \frac{8x^3}{3}+15e^{-x}$
I want to know if this result is correct guys
The result you obtained is correct. We elaborate some of the steps here. We have the differential equation:
$$ y'' + y' = 8x^2 $$
with $y(0) = 0$ and $y'(0) = 1$. Taking the Laplace transform we get:
$$ s^2Y(s) - 1 + sY(s) = \frac{16}{s^3}$$
We rewrite as:
$$ Y(s) = \frac{16 + s^3}{s^4(s + 1)} $$
Applying partial fraction decomposition we get:
$$ Y(s) = \frac{16}{s^4} - \frac{16}{s^3} + \frac{16}{s^2} + \frac{15}{s + 1} - \frac{15}{s} $$
Using the following two identities:
$$ L \left(x^n\right) = \frac{n!}{s^{n + 1}} $$ $$ L \left( e^{ax} \right) = \frac{1}{s - a}$$
we take the inverse transform and get:
$$ y(x) = \frac{8}{3} x^3 - 8x^2 + 16x + 15e^{-x} - 15 $$
which is the result you obtained.