Question

Find the Laplace Transform of y''+7y'+7y given that y(0)=0 and y'(0)=7

Answer 1

Assuming you start with the homogeneous ODE,

$y''+7y'+7y=0$

upon taking the Laplace transform of both sides, you end up with

$\mathcal L\left\{y''+7y'+7y\right\}=\mathcal L\{0\}$
$\mathcal L\{y''\}+7\mathcal L\{y'\}+7\mathcal L\{y\}=0$

since the transform operator is linear, and the transform of 0 is 0.

I'll denote the Laplace transform of a function $y(t)$ into the $s$-domain by $\mathcal L_s\{y(t)\}:=Y(s)$.

Given the derivative of $y(t)$, its Laplace transform can be found easily from the definition of the transform itself:

$Y(s)=\displaystyle\int_0^\infty y(t)e^{-st}\,\mathrm dt$
$\implies\mathcal L_s\{y'(t)\}=\displaystyle\int_0^\infty y'(t)e^{-st}\,\mathrm dt$

Integrate by parts, setting

$u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt$
$\mathrm dv=y'(t)\,\mathrm dt\implies v=y(t)$

so that

$\mathcal L_s\{y'(t)\}=y(t)e^{-st}\bigg|_{t=0}^{t\to\infty}+s\displaystyle\int_0^\infty y(t)e^{-st}\,\mathrm dt$

The second term is just the transform of the original function, while the first term reduces to $y(0)$ since $e^{-st}\to0$ as $t\to\infty$, and $e^{-st}\to1$ as $t\to0$. So we have a rule for transforming the first derivative, and by the same process we can generalize it to any order provided that we're given the value of all the preceeding derivatives at $t=0$.

The general rule gives us

$\mathcal L_s\{y(t)\}=Y(s)$
$\mathcal L_s\{y'(t)\}=sY(s)-y(0)$
$\mathcal L_s\{y''(t)\}=s^2Y(s)-sy(0)-y'(0)$

and so our ODE becomes

$\bigg(s^2Y(s)-sy(0)-y'(0)\bigg)+7\bigg(sY(s)-y(0)\bigg)+7Y(s)=0$
$(s^2+7s+7)Y(s)-7=0$
$Y(s)=\dfrac7{s^2+7s+7}$
$Y(s)=\dfrac{14}{\sqrt{21}}\dfrac{\frac{\sqrt{21}}2}{\left(s+\frac72\right)^2-\left(\frac{\sqrt{21}}2\right)^2}$

Depending on how you learned about finding inverse transforms, you should either be comfortable with cross-referencing a table and do some pattern-matching, or be able to set up and compute an appropriate contour integral. The former approach seems to be more common, so I'll stick to that.

Recall that

$\mathcal L_s\{\sinh(at)\}=\dfrac a{s^2-a^2}$

and that given a function $y(t)$ with transform $Y(s)$, the shifted transform $Y(s-c)$ corresponds to the function $e^{ct}y(t)$.

We have

$Y(s)=\dfrac{14}{\sqrt{21}}\dfrac{\frac{\sqrt{21}}2}{\left(s+\frac72\right)^2-\left(\frac{\sqrt{21}}2\right)^2}$
$\implies Y\left(s-\dfrac72\right)=\dfrac{14}{\sqrt{21}}\dfrac{\frac{\sqrt{21}}2}{s^2-\left(\frac{\sqrt{21}}2\right)^2}$

and so the inverse transform for our ODE is

$\mathcal L^{-1}_t\left\{Y\left(s-\dfrac72\right)\right\}=\dfrac{14}{\sqrt{21}}\mathcal L^{-1}_t\left\{\dfrac{\frac{\sqrt{21}}2}{s^2-\left(\frac{\sqrt{21}}2\right)^2}\right\}$
$\implies e^{7/2t}y(t)=\dfrac{14}{\sqrt{21}}\sinh\left(\dfrac{\sqrt{21}}2t\right)$
$\implies y(t)=\dfrac{14}{\sqrt{21}}e^{-7/2t}\sinh\left(\dfrac{\sqrt{21}}2t\right)$

and in case you're not familiar with hyperbolic functions, you have

$\sinh t=\dfrac{e^t-e^{-t}}2$