Q22E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the differential equation.$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{10}\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{10}{\mathbf{x}}^4\mathbf{+}\mathbf{24}{\mathbf{x}}^3\mathbf{+}\mathbf{2}{\mathbf{x}}^2\mathbf{-}\mathbf{12}\mathbf{x}\mathbf{+}\mathbf{18}$

Step-by-Step Solution

Answer

The general solution to the given differential equation is:

$\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{-3\mathrm{x}}\mathrm{cosx}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{x}}\mathrm{sinx}+{\mathrm{x}}^4-{\mathrm{x}}^2+2$.

Step 1: Write the auxiliary equation of the given differential equation.

The differential equation is,

 $\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+6\mathrm{y}\text{'}\left(\mathrm{x}\right)+10\mathrm{y}\left(\mathrm{x}\right)=10{\mathrm{x}}^4+24{\mathrm{x}}^3+2{\mathrm{x}}^2-12\mathrm{x}+18 \dots \left(1\right)$

 

Write the homogeneous differential equation of the equation (1),

 

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+6\mathrm{y}\text{'}\left(\mathrm{x}\right)+10\mathrm{y}\left(\mathrm{x}\right)=0$

 

The auxiliary equation for the above equation,

 

${\mathrm{m}}^2+6\mathrm{m}+10=0$

 

Step 2: Find the complementary solution of the given equation.

Solve the auxiliary equation,

 $\begin{array}{c}{\mathrm{m}}^2+6\mathrm{m}+10=0\\ \mathrm{m}=\frac{-6\pm \sqrt{36-40}}{2}\\ \mathrm{m}=\frac{-6\pm \sqrt{4}}{2}\\ \mathrm{m}=-3\pm \mathrm{i}\end{array}$

 

The roots of the auxiliary equation are, 

 

${\mathrm{m}}_1=-3+\mathrm{i},{\mathrm{m}}_2=-3-\mathrm{i}$

 

The complementary solution of the given equation is,

 ${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{-3\mathrm{x}}\mathrm{cosx}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{x}}\mathrm{sinx}$

 

Step 3: Now find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{Ax}}^4+{\mathrm{Bx}}^3+{\mathrm{Cx}}^2+\mathrm{Dx}+\mathrm{E}......\left(2\right)$

Now find the first and second derivatives of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=4{\mathrm{Ax}}^3+3{\mathrm{Bx}}^2+2\mathrm{Cx}+\mathrm{D}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=12{\mathrm{Ax}}^2+6\mathrm{Bx}+2\mathrm{C}\end{array}$

Substitute the value of  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$ the equation (1),

$\begin{array}{l}\Rightarrow \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)+6\mathrm{y}\text{'}\left(\mathrm{x}\right)+10\mathrm{y}\left(\mathrm{x}\right)=10{\mathrm{x}}^4+24{\mathrm{x}}^3+2{\mathrm{x}}^2-12\mathrm{x}+18\\ \Rightarrow 12{\mathrm{Ax}}^2+6\mathrm{Bx}+2\mathrm{C}+6(4{\mathrm{Ax}}^3+3{\mathrm{Bx}}^2+2\mathrm{Cx}+\mathrm{D})+10({\mathrm{Ax}}^4+{\mathrm{Bx}}^3+{\mathrm{Cx}}^2+\mathrm{Dx}+\mathrm{E})\\ =10{\mathrm{x}}^4+24{\mathrm{x}}^3+2{\mathrm{x}}^2-12\mathrm{x}+18\\ \Rightarrow 10{\mathrm{Ax}}^4+(10\mathrm{B}+24\mathrm{A}){\mathrm{x}}^3+(12\mathrm{A}+18\mathrm{B}+10\mathrm{C}){\mathrm{x}}^2+(6\mathrm{B}+12\mathrm{C}+10\mathrm{D})\mathrm{x}+(2\mathrm{C}+6\mathrm{D}+10\mathrm{E})\\ =10{\mathrm{x}}^4+24{\mathrm{x}}^3+2{\mathrm{x}}^2-12\mathrm{x}+18\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}10\mathrm{A}=10\Rightarrow \mathrm{A}=1\\ 10\mathrm{B}+24\mathrm{A}=24 \dots \left(3\right)\\ 12\mathrm{A}+18\mathrm{B}+10\mathrm{C}=2 \dots \left(4\right)\\ 6\mathrm{B}+12\mathrm{C}+10\mathrm{D}=-12 \dots \left(5\right)\\ 2\mathrm{C}+6\mathrm{D}+10\mathrm{E}=18\dots \left(6\right)\end{array}$

Substitute the value of A in the equation (3),

$\begin{array}{c}10\mathrm{B}+24\left(1\right)=24\\ \mathrm{B}=0\end{array}$

Substitute the value of A and B in the equation (4),

$\begin{array}{c}12\left(1\right)+18\left(0\right)+10\mathrm{C}=2\\ \mathrm{C}=-1\end{array}$

Substitute the value of C and B in the equation (5),

$\begin{array}{c}6\left(0\right)+12(-1)+10\mathrm{D}=-12\\ \mathrm{D}=0\end{array}$

Substitute the value of C and D in the equation (6),

$\begin{array}{c}2(-1)+6\left(0\right)+10\mathrm{E}=18\\ \mathrm{E}=2\end{array}$

Substitute the value of A, B, C, D, and E in the equation (2),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{Ax}}^4+{\mathrm{Bx}}^3+{\mathrm{Cx}}^2+\mathrm{Dx}+\mathrm{E}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\left(1\right){\mathrm{x}}^4+\left(0\right){\mathrm{x}}^3+(-1){\mathrm{x}}^2+\left(0\right)\mathrm{x}+\left(2\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{x}}^4-{\mathrm{x}}^2+2\end{array}$

Therefore, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{x}}^4-{\mathrm{x}}^2+2$

Step 4: Conclusion

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{-3\mathrm{x}}\mathrm{cosx}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{x}}\mathrm{sinx}+{\mathrm{x}}^4-{\mathrm{x}}^2+2\end{array}$