Question

(t)= . Use the Laplace transform to solve the following initial value problem: 44" + 2y + 18y = 3 cos(3+), y(0) = 0, y(0) = 0. a. First, take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation and then solve for L{y(t)}. Do not perform partial fraction decomposition since we will write the solution in terms of a convolution integral. L{y(t)}(s) b. Express the solution y(t) in terms of a convolution integral. dw

Answer 1

We have,

$4y''+2y'+18y=3\cos(3t),y(0)=0,y'(0)=0$

Taking Laplace transform both sides we get:

L {4y' + 2y + 18y} = L {3 cos(3t)}

→ 4L {y'} + 2L {y'} + 18L {y} = 3L {cos(3t)}

→ 4(-L {y} – sy(O)- y0))+2(SL {y} - y0)+18L {y} = 3L {cos(3t)}

:: Ly" = sºL {y} – sy(0) - y0), L{y} = sl {y} - y0)

$\Rightarrow 4\left( s^2L\left \{ y \right \}-s.0-0\right )+2\left ( sL\left \{ y \right \}-0 \right )+18L\left \{ y\right \}=3L\left \{ \cos(3t) \right \}$

$\left [ \because y(0)=0,y'(0)=0 \right ]$

$\Rightarrow 4 s^2L\left \{ y \right \}+2sL\left \{ y \right \}+18L\left \{ y\right \}=3L\left \{ \cos(3t) \right \}$

$\Rightarrow (4 s^2+2s+18)L\left \{ y \right \}=3L\left \{ \cos(3t) \right \}$

$\Rightarrow (4 s^2+2s+18)L\left \{ y \right \}=3\frac{s}{s^2+9}$

$\Rightarrow {\color{Blue} L\left \{ y(t) \right \}(s)=\frac{3s}{(4 s^2+2s+18)(s^2+9)}}$

$=\frac{3}{4}\frac{1}{(s+\frac{1}{4})^2+\frac{71}{16}}.\frac{s}{s^2+9}$

$=\frac{3}{4}L\left \{ e^{-\frac{1}{4}t}.\sin(\frac{\sqrt{71}}{4}t) \right \}.L\left \{ \cos(3t) \right \}$

$=\frac{3}{4}L\left \{ \int_{0}^{t}\cos3(t-\tau ).e^{-\frac{1}{4}\tau }.\sin(\frac{\sqrt{71}}{4}\tau )d\tau \right \}$

$=L\left \{ \frac{3}{4}\int_{0}^{t}\cos(3t-3\tau ).e^{-\frac{1}{4}\tau }.\sin(\frac{\sqrt{71}}{4}\tau )d\tau \right \}$

$\Rightarrow {\color{Blue} y(t)=\frac{3}{4} \int_{0}^{t}\cos(3t-3\tau ).e^{-\frac{1}{4}\tau }.\sin(\frac{\sqrt{71}}{4}\tau )d\tau}$

  ${\color{Blue} = \int_{0}^{t}\frac{3}{4}\cos(3t-3\tau ).e^{-\frac{1}{4}\tau }.\sin(\frac{\sqrt{71}}{4}\tau )d\tau}$

  ${\color{Blue} = \int_{0}^{t} \frac{3}{4} e^{-\frac{1}{4}(t-\tau)} \sin(\frac{\sqrt{71}}{4}(t-\tau)) \cos(3\tau ) d\tau}$

NOTE:

• $L^{-1}\left \{ \frac{1}{(s+a)^2+b^2} \right \}=e^{-at}.\sin(bt)$
• $L\left \{ f(t) \right \}.L\left \{ g(t) \right \}=L\left \{ \int_{0}^{t}f(t-\tau)g(\tau)d\tau \right \}$, convolution theorem

Then, we get

$\frac{3s}{(4 s^2+2s+18)(s^2+9)}$$= \frac{3}{4}L\left \{\int_{0}^{t}\cos3(t-\tau ).e^{-\frac{1}{4}\tau }.\sin(\frac{\sqrt{71}}{4}\tau )d\tau \right \}$

Hope this helps!