Solve the inital value problem by using Laplace transform:

${y}^{\u2033}-5{y}^{\prime}+6y=-8\mathrm{cos}(t)-2\mathrm{sin}(t),y(\frac{\pi}{2})=1,{y}^{\prime}(\frac{\pi}{2})=0$

Amari Flowers
2020-11-09
Answered

Solve the inital value problem by using Laplace transform:

${y}^{\u2033}-5{y}^{\prime}+6y=-8\mathrm{cos}(t)-2\mathrm{sin}(t),y(\frac{\pi}{2})=1,{y}^{\prime}(\frac{\pi}{2})=0$

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SchulzD

Answered 2020-11-10
Author has **83** answers

Solve the inital value problem by using Laplace transform

${y}^{\u2033}-5{y}^{\prime}+6y=-8\mathrm{cos}(t)-2\mathrm{sin}(t),y(\frac{\pi}{2})=1,{y}^{\prime}(\frac{\pi}{2})=0$

Step 2

Consider:${y}^{\u2033}-5{y}^{\prime}+6y=-8\mathrm{cos}(t)-2\mathrm{sin}(t),y(\frac{\pi}{2})=1,{y}^{\prime}(\frac{\pi}{2})=0$ 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$t=\eta +\frac{\pi}{2}\Rightarrow \eta =t-\frac{\pi}{2}$

${y}^{\u2033}(\eta +\frac{\pi}{2})-5{y}^{\prime}(\eta +\frac{\pi}{2})+6y(\eta +\frac{\pi}{2})=-8\mathrm{cos}(\eta +\frac{\pi}{2})-2\mathrm{sin}(\eta +\frac{\pi}{2})$

${y}^{\u2033}(\eta +\frac{\pi}{2})-5{y}^{\prime}(\eta +\frac{\pi}{2})+6y(\eta +\frac{\pi}{2})=8\mathrm{sin}(\eta )-2\mathrm{cos}(\eta )\dots (2)$

We have,$u(\eta )=y(\eta +\frac{\pi}{2})$

By using chain rule,

${u}^{\prime}(\eta )=\frac{du}{d\eta}=\frac{dy}{d\eta}\cdot \frac{d\eta}{dt}={y}^{\prime}(\eta +\frac{\pi}{2})$

${u}^{\prime}(\eta )=\frac{du}{d\eta}=\frac{dy}{d\eta}\cdot \frac{d\eta}{dt}={y}^{\prime}(\eta +\frac{\pi}{2})$

Hence,the initial condition for$u(\eta )$ as

$u(0)=y\left(\frac{\pi}{2}\right)=1\text{and}{u}^{\prime}(0)={y}^{\prime}\left(\frac{\pi}{2}\right)=0$

Step 3

Now,the equation (2) becomes

${u}^{\u2033}(\eta )-5{u}^{\prime}(\eta )+6u(\eta )=8\mathrm{sin}(\eta )-2\mathrm{cos}(\eta )$

Apply the Laplace transform on both sides,then

$L\{{u}^{\u2033}(\eta )\}-5L\{{u}^{\prime}(\eta )\}+6L\{u(\eta )\}=8L\{\mathrm{sin}(\eta )\}-2L\{\mathrm{cos}(\eta )\}$

${s}^{2}u(s)-su(0)-{u}^{\prime}(0)-5su(s)-u(0)+6u(s)=\frac{8}{({s}^{2}+1)}-\frac{2s}{({s}^{2}+1)}$

${s}^{2}u(s)-s-5su(s)+5+6u(s)=\frac{8}{({s}^{2}+1)}-\frac{2s}{({s}^{2}+1)}$

$({s}^{2}+1)u(s)=\frac{8}{({s}^{2}+1)}-\frac{2s}{({s}^{2}+1)}+s-5$

Step 2

Consider:

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

We have,

By using chain rule,

Hence,the initial condition for

Step 3

Now,the equation (2) becomes

Apply the Laplace transform on both sides,then