 # 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 Amari Flowers 2020-11-09 Answered
Solve the inital value problem by using Laplace transform:
${y}^{″}-5{y}^{\prime }+6y=-8\mathrm{cos}\left(t\right)-2\mathrm{sin}\left(t\right),y\left(\frac{\pi }{2}\right)=1,{y}^{\prime }\left(\frac{\pi }{2}\right)=0$
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Solve the inital value problem by using Laplace transform
${y}^{″}-5{y}^{\prime }+6y=-8\mathrm{cos}\left(t\right)-2\mathrm{sin}\left(t\right),y\left(\frac{\pi }{2}\right)=1,{y}^{\prime }\left(\frac{\pi }{2}\right)=0$
Step 2
Consider:${y}^{″}-5{y}^{\prime }+6y=-8\mathrm{cos}\left(t\right)-2\mathrm{sin}\left(t\right),y\left(\frac{\pi }{2}\right)=1,{y}^{\prime }\left(\frac{\pi }{2}\right)=0$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 $t=\eta +\frac{\pi }{2}⇒\eta =t-\frac{\pi }{2}$
${y}^{″}\left(\eta +\frac{\pi }{2}\right)-5{y}^{\prime }\left(\eta +\frac{\pi }{2}\right)+6y\left(\eta +\frac{\pi }{2}\right)=-8\mathrm{cos}\left(\eta +\frac{\pi }{2}\right)-2\mathrm{sin}\left(\eta +\frac{\pi }{2}\right)$
${y}^{″}\left(\eta +\frac{\pi }{2}\right)-5{y}^{\prime }\left(\eta +\frac{\pi }{2}\right)+6y\left(\eta +\frac{\pi }{2}\right)=8\mathrm{sin}\left(\eta \right)-2\mathrm{cos}\left(\eta \right)\dots \left(2\right)$
We have, $u\left(\eta \right)=y\left(\eta +\frac{\pi }{2}\right)$
By using chain rule,
${u}^{\prime }\left(\eta \right)=\frac{du}{d\eta }=\frac{dy}{d\eta }\cdot \frac{d\eta }{dt}={y}^{\prime }\left(\eta +\frac{\pi }{2}\right)$
${u}^{\prime }\left(\eta \right)=\frac{du}{d\eta }=\frac{dy}{d\eta }\cdot \frac{d\eta }{dt}={y}^{\prime }\left(\eta +\frac{\pi }{2}\right)$
Hence,the initial condition for $u\left(\eta \right)$ as

Step 3
Now,the equation (2) becomes
${u}^{″}\left(\eta \right)-5{u}^{\prime }\left(\eta \right)+6u\left(\eta \right)=8\mathrm{sin}\left(\eta \right)-2\mathrm{cos}\left(\eta \right)$
Apply the Laplace transform on both sides,then
$L\left\{{u}^{″}\left(\eta \right)\right\}-5L\left\{{u}^{\prime }\left(\eta \right)\right\}+6L\left\{u\left(\eta \right)\right\}=8L\left\{\mathrm{sin}\left(\eta \right)\right\}-2L\left\{\mathrm{cos}\left(\eta \right)\right\}$
${s}^{2}u\left(s\right)-su\left(0\right)-{u}^{\prime }\left(0\right)-5su\left(s\right)-u\left(0\right)+6u\left(s\right)=\frac{8}{\left({s}^{2}+1\right)}-\frac{2s}{\left({s}^{2}+1\right)}$
${s}^{2}u\left(s\right)-s-5su\left(s\right)+5+6u\left(s\right)=\frac{8}{\left({s}^{2}+1\right)}-\frac{2s}{\left({s}^{2}+1\right)}$
$\left({s}^{2}+1\right)u\left(s\right)=\frac{8}{\left({s}^{2}+1\right)}-\frac{2s}{\left({s}^{2}+1\right)}+s-5$