Use Laplace transformations to solve $y''+4y'+4y=e^{-x}$, so that $y(0)=0$ and $y'(0)=1$

Question

Use Laplace transformations to solve $y''+4y'+4y=e^{-x}$, so that $y(0)=0$ and $y'(0)=1$.

I applied the transformation but I don't understand the rest of the process. Can anyone explain me based on this exercise?

Thanks

Answer 1

Laplace transforms are great because they are just algebra intensive. The process begins by transforming your equation. $$\mathcal{L}\{y''+4y'+4y=e^{-x} \}\\ \implies s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+4Y(s) = \frac{1}{s+1}$$ Now plug in $y(0)= 0, \space y'(0)=1$ to get $$s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+4Y(s)=s^2Y(s)-1+4sY(s)+4Y(s)$$ Factor out $Y(s)$ on the RHS of the equation above to get $$Y(s)(s^2+4s+4)-1=\frac{1}{1+s} \\ \implies Y(s)(s^2+4s+4) = \frac{s+2}{s+1} \\ \implies Y(s) = \frac{s+2}{(s+1)(s^2+4s+4)}$$ Now simply apply an inverse transform to both sides and you'll get back $y(x)$. The quantity $ \frac{s+2}{(s+1)(s^2+4s+4)}$ may look challenging to inverse transform. Remember that algebra is your friend in Laplace transforms, and observing that $(s^2+4s+4)$ can easily be factored into a square will make things a lot easier. Lastly, you should be able to use partial fraction decomposition, at which point you'll be left with an expression that should be very easy to inverse transform.

Answer 2

The process is:

• Apply Laplace transform, this yields an equation that has some terms with $\mathscr L\{y(x)\}(s)$.

• Solve for $\mathscr L\{y(x)\}(s)$, which is normally denoted by $Y(s)$.

• Once you solved the equation, apply the inverse of laplace transform and you will get $y(x)$.

Example:

$$y'+4y=1,~~~~~~~~~~~ y(0)=1.$$

If we apply Laplace transform:

$$sY(s)-y(0)+4Y(s)=\frac{1}{s},$$ $$Y(s)(s+4)=\frac{1}{s}+1.$$ Now we solve for $Y(s)$: $$Y(s)=\frac{1}{s(s+4)}+\frac{1}{s+4}\stackrel{\text{Use partial fractions.}}{=}\frac{1}{4s}+\frac{3}{4(s+4)}.$$ Finally we apply inverse Laplace transform: $$y(x)=\frac{1}{4}+\frac{3}{4}e^{-4x}.$$