Q19E - 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 non-homogeneous equation. 

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

$$\begin{gathered} \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{5}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{+}\mathbf{6}\mathbf{x}\mathbf{-}\mathbf{5}{\mathbf{x}}^{\mathbf{2}}\mathbf{;} \\ \mathbf{y}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{;} \\ {\mathbf{y}}_{\mathbf{p}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}\mathbf{;}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\left\{e^x, e^{-x}\cos \left(2x\right), e^{-x}\sin \left(2x\right)\right\} \end{gathered}$$

Step-by-Step Solution

Answer

(a) The value is $y=c_1{e}^x+c_2{e}^{-x}\cos \left(2x\right)+c_3{e}^{-x}\sin \left(2x\right)+x^2$

(b) The value is $y=e^{-x}\sin \left(2x\right)-e^x+x^2$

(a) Step 1: Firstly solve for y n

The given equation is, $y\text{'}\text{'}\text{'}+y\text{'}\text{'}+3y\text{'}-5y=2+6x-5x^2$

 

Solve for, $y_n$

 

$y\text{'}\text{'}\text{'}+y\text{'}\text{'}+3y\text{'}-5y=0$

 

Solve the above equation,

 $$\begin{gathered} D^3+D^2+3D-5=0 \\ {D}^3+2D^2-D^2+5D-2D-5=0 \\ {D}^3+2D^2+5D-D^2-2D-5=0 \\ D\left(D^2+2D+5\right)-1\left(D^2+2D+5\right)=0 \end{gathered}$$

$$\begin{gathered} \left(D-1\right)\left(D^2+2D+5\right)=0 \\ D-1=0 or D^2+2D+5=0 \\ D=1, D=-1+2i, D=-1-2i \end{gathered}$$

 

Then, the general solution is $y_n=c_1{e}^x+c_2{e}^{-x}\cos \left(2x\right)+c_3{e}^{-x}\sin \left(2x\right)$

Step 2: A general solution to the nonhomogeneous equation

$$\begin{gathered} y=y_n+y_p \\ y=c_1{e}^x+c_2{e}^{-x}\cos \left(2x\right)+c_3{e}^{-x}\sin \left(2x\right)+x^2 \end{gathered}$$

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

Given initial conditions are, $y\left(0\right)=-1, y\text{'}\left(0\right)=1, y\text{'}\text{'}\left(0\right)=-3;$

 

Firstly, solve for, $y\left(0\right)=-1$

 

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

 

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

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

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

One has,

 

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

 

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

 

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

Step 5: Now, solve for , y '' (0)=-3

One has, 

 $$\begin{gathered} y\text{'}=c_1{e}^x+c_2\left(-e^{-x}\cos \left(2x\right)-2e^{-x}\sin \left(2x\right)\right)+c_3\left(-e^{-x}\sin \left(2x\right)+2e^{-x}\cos \left(2x\right)\right)+2x\backslash \mathrm{hfill} \\ y\text{'}\text{'}=c_1{e}^x+c_2\left(-3e^{-x}\cos \left(2x\right)+4e^{-x}\sin \left(2x\right)\right)+c_3\left(-3e^{-x}\sin \left(2x\right)-4e^{-x}\cos \left(2x\right)\right)+2\backslash \mathrm{hfill} \end{gathered}$$

 

Substitute $y\text{'}\text{'}\left(0\right)=-3$ in the above equation,

 

$$\begin{gathered} -3=c_1{e}^0+c_2\left(-3e^{-\left(0\right)}\cos \left(2\left(0\right)\right)+4e^{-\left(0\right)}\sin \left(2\left(0\right)\right)\right) \\ +c_3\left(-3e^{-\left(0\right)}\sin \left(2\left(0\right)\right)-4e^{-\left(0\right)}\cos \left(2\left(0\right)\right)\right)+2 \\ -3=c_1+c_2\left(-3\right)+c_3\left(-4\right)+2 \\ {c}_1-3c_2-4c_3+2=-3 \\ {c}_1-3c_2-4c_3=-5 ......\mathbf{\left(}\mathbf{3}\mathbf{\right)} \end{gathered}$$

Step 6: find the value of c 1 , c 2 and , c 3

Solve the equation (2) and (3),

 

$$\begin{gathered} 2\left(c_1-c_2+2c_3\right)=1\times 2 \\ 2c_1-2c_2+4c_3=2 \\ {c}_1-3c_2-4c_3=-5 \\ 3c_1-5c_2=-3= ......\mathbf{\left(}\mathbf{4}\mathbf{\right)} \end{gathered}$$

 

Now, solve the equation (1) and (4),

 $$\begin{gathered} \left(c_1+c_2=-1\right)5 \\ 5c_1+5c_2=-5 \\ 3c_1-5c_2=-3 \\ \overline{8c_1=-8} \\ c_1=-1 \end{gathered}$$

 

Substitute the value of $c_1$  in the equation (1),

$$\begin{gathered} c_1+c_2=-1 \\ \left(-1\right)+c_2=-1 \\ {c}_2=0 \end{gathered}$$

 

Substitute the value of $c_1, c_2$  in the equation (2),

$$\begin{gathered} c_1-c_2+2c_3=1 \\ -1-0+2c_3=1 \\ 2c_3=2 \\ {c}_3=1 \end{gathered}$$

Step 7: Final conclusion, the solution that satisfies the specified initial condition.

Substitute the value of $c_1, c_2$ and $c_3$ in the general solution.

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