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

Question
Laplace transform
asked 2020-11-09
Solve the inital value problem by using Laplace transform:
\(y''-5y'+6y=-8\cos(t)-2\sin(t), y(\frac{\pi}{2})=1 ,y'(\frac{\pi}{2})=0\)

Answers (1)

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, \(y(t)=(t-\frac{\pi}{2})\left[-4\sin(t)-\cos(t)\right]+3\cos(t)+\sin(t)\)
0

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