Question

# Solve the inital value problem by using Laplace transform: y''-5y'+6y=-8cos(t)-2sin(t), y(frac{pi}{2})=1 ,y'(frac{pi}{2})=0

Laplace transform
Solve the inital value problem by using Laplace transform:
$$y''-5y'+6y=-8\cos(t)-2\sin(t), y(\frac{\pi}{2})=1 ,y'(\frac{\pi}{2})=0$$

2020-11-10
Solve the inital value problem by using Laplace transform
$$y''-5y'+6y=-8\cos(t)-2\sin(t), y(\frac{\pi}{2})=1 ,y'(\frac{\pi}{2})=0$$
Step 2
Consider:$$y''-5y'+6y=-8\cos(t)-2\sin(t), y(\frac{\pi}{2})=1 ,y'(\frac{\pi}{2})=0$$First noticed that the initial condition are not at t=0.We can applythe Laplace transform of derivative is to have the initial condition at t=0.
This mean that we will need to formulate the IVP in such a way thatthe initial condition are at t=0.
We need to change the variable $$t=\eta+\frac{\pi}{2}\Rightarrow\eta=t-\frac{\pi}{2}$$
$$y''(\eta+\frac{\pi}{2})-5y'(\eta+\frac{\pi}{2})+6y(\eta+\frac{\pi}{2})=-8\cos(\eta+\frac{\pi}{2})-2\sin(\eta+\frac{\pi}{2})$$
$$y''(\eta+\frac{\pi}{2})−5y'(\eta+\frac{\pi}{2})+6y(\eta+\frac{\pi}{2})=8\sin(\eta)-2\cos(\eta)\dots(2)$$
We have, $$u(\eta)=y(\eta+\frac{\pi}{2})$$
By using chain rule,
$$u'(\eta)=\frac{du}{d \eta}=\frac{dy}{d \eta}\cdot \frac{d \eta}{dt}=y'\left(\eta+\frac{\pi}{2}\right)$$
$$u'(\eta)=\frac{du}{d \eta}=\frac{dy}{d \eta}\cdot \frac{d \eta}{dt}=y'\left(\eta+\frac{\pi}{2}\right)$$
Hence,the initial condition for $$u(\eta)$$ as
$$u(0)=y\left(\frac{\pi}{2}\right)=1 \text{ and } u'(0)=y'\left(\frac{\pi}{2}\right)=0$$
Step 3
Now,the equation (2) becomes
$$u''(\eta)-5u'(\eta)+6u(\eta)=8\sin(\eta)-2\cos(\eta)$$
Apply the Laplace transform on both sides,then
$$L\left\{u''(\eta)\right\}-5L\left\{u'(\eta)\right\}+6L\left\{u(\eta)\right\}=8L\left\{\sin(\eta)\right\}-2L\left\{\cos(\eta)\right\}$$
$$s^2u(s)-su(0)-u'(0)-5{su(s)-u(0)}+6u(s)=\frac{8}{(s^2+1)}-\frac{2s}{(s^2+1)}$$
$$s^2u(s)-s-5su(s)+5+6u(s)=\frac{8}{(s^2+1)}-\frac{2s}{(s^2+1)}$$
$$(s^2+1)u(s)=\frac{8}{(s^2+1)}-\frac{2s}{(s^2+1)}+s-5$$
$$\Rightarrow u(s)=\frac{8}{(s^2+1)^2}-\frac{2s}{(s^2+1)^2}+\frac{s}{(s^2+1)}-\frac{5}{(s^2+1)}$$
Step 4
Apply the Laplace inverse transform on the both sides,then
$$L^{-1}\left\{u(s)\right\}=8L^{-1}\left\{\frac{1}{(s^2+1)^2}\right\}-2L^{-1}\left\{\frac{s}{(s^2+1)^2}\right\}+L^{-1}\left\{\frac{s}{s^2+1}\right\}-L^{-1}\left\{\frac{1}{s^2+1}\right\}$$
$$u(t)=8\left\{\frac{1}{2}(-\eta \cos(\eta)+\sin(\eta))\right\}-2\left\{\frac{\eta\sin(\eta)}{2}\right\}+\cos(\eta)-\sin(\eta)$$
$$u(t)=-4\eta \cos(\eta)+4\sin(\eta)- \eta \sin(\eta)+\cos(\eta)-\sin(\eta)$$
$$\Rightarrow u(t)=-4\eta \cos(\eta)+3\sin(\eta)- \eta \sin(\eta)+\cos(\eta)$$ Now,put $$\eta=t-\frac{\pi}{2}$$, then
$$u(t-\frac{\pi}{2})=-4(t-\frac{\pi}{2})\cos(t-\frac{\pi}{2})+3\sin(t-\frac{\pi}{2})-(t-\frac{\pi}{2})\sin(t-\frac{\pi}{2})+\cos(t-\frac{\pi}{2})$$
$$y(t)=-4(t-\frac{\pi}{2})\sin(t)+3\cos(t)-(t-\frac{\pi}{2})\cos(t)+\sin(t)$$
Hence, $$y(t)=(t-\frac{\pi}{2})\left[-4\sin(t)-\cos(t)\right]+3\cos(t)+\sin(t)$$