Q22E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A particular solution and a fundamental solution set are given for a nonhomogeneous equation and its corresponding homogeneous equation. 

(a) Find a general solution to the nonhomogeneous equation. 

(b) Find the solution that satisfies the specified initial condition.

$$\begin{gathered} {\mathbf{y}}^{\left(4\right)}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{5}\mathbf{cosx}\mathbf{;} \\ \mathbf{y}\left(0\right)\mathbf{=}\mathbf{2}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{;} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{=}\mathbf{cosx}\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\left\{e^x\mathrm{cosx}, e^x\mathrm{sinx}, e^{-x}\mathrm{cosx}, e^{-x}\mathrm{sinx}\right\} \end{gathered}$$

Step-by-Step Solution

Answer

(a) The value is $y=c_1{e}^x\cos x+c_2{e}^x\sin x+c_3{e}^{-x}\cos x+c_4{e}^{-x}\sin x+\cos x$

(b) The value is $y=e^x\cos x+\cos x$

(a) Step 1: Firstly solve for y n

The given equation is, $y^{\left(4\right)}-4y=5\cos x$

Solve for $y_n$,

$y^{\left(4\right)}-4y=0$

Here it is given that; the fundamental solution set for the homogeneous equation is,

$\left\{e^x\cos x, e^x\sin x, e^{-x}\cos x, e^{-x}\sin x\right\}$

Then, the general solution is:

$y_n=c_1{e}^x\cos x+c_2{e}^x\sin x+c_3{e}^{-x}\cos x+c_4{e}^{-x}\sin x$

Step 2 :A general solution to the nonhomogeneous equation.

$$\begin{gathered} y=y_n+y_p \\ y=c_1{e}^x\cos x+c_2{e}^x\sin x+c_3{e}^{-x}\cos x+c_4{e}^{-x}\sin x+\cos x \end{gathered}$$

(b) Step 3:Solve for given initial conditions.

Given initial conditions are, $y\left(0\right)=2, y\text{'}\left(0\right)=1, y\text{'}\text{'}\left(0\right)=-1, y\text{'}\text{'}\text{'}\left(0\right)=-2;$

Firstly, solve for, $y\left(0\right)=2$

We have, 

$y=c_1{e}^x\cos x+c_2{e}^x\sin x+c_3{e}^{-x}\cos x+c_4{e}^{-x}\sin x+\cos x$

Substitute $y\left(0\right)=2$ in the above equation

,$$\begin{gathered} 2=c_1{e}^{\left(0\right)}\cos \left(0\right)+c_2{e}^{\left(0\right)}\sin \left(0\right)+c_3{e}^{-\left(0\right)}\cos \left(0\right)+c_4{e}^{-\left(0\right)}\sin \left(0\right)+\cos \left(0\right) \\ 2=c_1+c_3+1 \\ {c}_1+c_3=1 ......\left(\mathbf{1}\right) \end{gathered}$$

Step 4: Now, solve for y ' (0)=1 ,

One has, 

$$\begin{gathered} y\text{'}=c_1\left(e^x\cos \left(x\right)-e^x\sin \left(x\right)\right)+c_2\left(e^x\sin \left(x\right)+\cos \left(x\right)e^x\right) \\ +c_3\left(-e^{-x}\cos \left(x\right)-e^{-x}\sin \left(x\right)\right)+c_4\left(-e^{-x}\sin \left(x\right)+e^{-x}\cos \left(x\right)\right)-\sin x \end{gathered}$$

Substitute $y\text{'}\left(0\right)=1$ in the above equation,

$$\begin{gathered} 1=c_1\left(e^0\cos \left(0\right)-e^0\sin \left(0\right)\right)+c_2\left(e^0\sin \left(0\right)+\cos \left(0\right)e^0\right) \\ +c_3\left(-e^{-\left(0\right)}\cos \left(0\right)-e^{-\left(0\right)}\sin \left(0\right)\right)+c_4\left(-e^{-\left(0\right)}\sin \left(0\right)+e^{-\left(0\right)}\cos \left(0\right)\right)-\sin \left(0\right) \\ 1=c_1+c_2+c_3\left(-1\right)+c_4 \\ {c}_1+c_2-c_3+c_4=1 ......\left(\mathbf{2}\right) \end{gathered}$$

Step 5: Now, solve for y '' (0)=-1 ,

One has, $y\text{'}\text{'}=-2c_1{e}^x\sin \left(x\right)+2c_2{e}^x\cos \left(x\right)+2c_3{e}^{-x}\sin \left(x\right)-2c_4{e}^{-x}\cos \left(x\right)-\cos \left(x\right)$

Substitute $y\text{'}\text{'}\left(0\right)=-1$ in the above equation,

$$\begin{gathered} -1=-2c_1{e}^0\sin \left(0\right)+2c_2{e}^0\cos \left(0\right)+2c_3{e}^{-\left(0\right)}\sin \left(0\right)-2c_4{e}^{-\left(0\right)}\cos \left(0\right)-\cos \left(0\right) \\ -1=2c_2-2c_4-1 \\ 2c_2-2c_4=0 ......\left(\mathbf{3}\right) \end{gathered}$$

Step 6: Now, solve for y'''(0)=-2 ,

One has,

$$\begin{gathered} y\text{'}\text{'}\text{'}=-2c_1\left(e^x\sin \left(x\right)+e^x\cos \left(x\right)\right)+2c_2\left(e^x\cos \left(x\right)-e^x\sin \left(x\right)\right)\backslash \mathrm{hfill} \\ +2c_3\left(-e^{-x}\sin \left(x\right)+e^{-x}\cos \left(x\right)\right)-2c_4\left(-e^{-x}\cos \left(x\right)-e^{-x}\sin \left(x\right)\right)+\sin \left(x\right)\backslash \mathrm{hfill} \end{gathered}$$

Substitute in the above equation, 

$$\begin{gathered} -2=-2c_1\left(e^0\sin \left(0\right)+e^0\cos \left(0\right)\right)+2c_2\left(e^0\cos \left(0\right)-e^0\sin \left(0\right)\right) \\ +2c_3\left(-e^{-0}\sin \left(0\right)+e^{-0}\cos \left(0\right)\right)-2c_4\left(-e^{-0}\cos \left(0\right)-e^{-0}\sin \left(0\right)\right)+\sin \left(0\right) \\ -2=-2c_1+2c_2+2c_3+2c_4 \\ -2c_1+2c_2+2c_3+2c_4=-2 ......\left(\mathbf{4}\right) \end{gathered}$$

Step 7: Find the value of c 1 ,c 2 , c 3 and , c 4

Solve the equation (2) and (4),

Solve the equation (3) and (5),

$$\begin{gathered} \left(2c_2-2c_4=0\right)\times 2 \\ 4c_2-4c_4=0 \\ 4c_2+4c_4=0 \\ \overline{8c_2=0 } \\ c_2=0 \end{gathered}$$

Substitute the value of  $c_2$ in the equation (3),

$$\begin{gathered} 2c_2-2c_4=0 \\ -2c_4=0 \\ {c}_4=0 \end{gathered}$$

Substitute the value of $c_2$ and $c_4$ in the equation (2),

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Solve the equation (1) and (6),

Substitute the value of   in the equation (1),

$$\begin{gathered} c_1+c_2-c_3+c_4=1 \\ {c}_1-c_3=1\dots \left(6\right) \end{gathered}$$