Q23E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{-}\mathbf{7}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{=}{\mathbf{\theta }}^2$

Step-by-Step Solution

Answer

The particular solution of the differential equation is ${\mathbf{y}}_p\mathbf{=}\mathbf{-}\frac{1}{21}{\mathbf{\theta }}^3\mathbf{-}\frac{1}{49}{\mathbf{\theta }}^2\mathbf{-}\frac{2}{343}\mathbf{\theta }$

Step 1: Firstly, write the auxiliary equation of the given differential equation

The differential equation is $\mathrm{y}\text{'}\text{'}\left(\mathrm{\theta }\right)-7\mathrm{y}\text{'}\left(\mathrm{\theta }\right)={\mathrm{\theta }}^2\dots \left(1\right)$

 

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

 $\mathrm{y}\text{'}\text{'}\left(\mathrm{\theta }\right)-7\mathrm{y}\text{'}\left(\mathrm{\theta }\right)=0$

 

The auxiliary equation for the above equation,

${\mathrm{m}}^2-7\mathrm{m}=0$

Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

 

$\begin{array}{c}{\mathrm{m}}^2-7\mathrm{m}=0\\ \mathrm{m}(\mathrm{m}-7)=0\end{array}$

 

The roots of the auxiliary equation are, 

 

${\mathrm{m}}_1=0,\&{\mathrm{m}}_2=7$

 

The complementary solution of the given equation is,

 

$\begin{array}{l}{\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{\left(0\right)\mathrm{\theta }}+{\mathrm{c}}_2{\mathrm{e}}^{7\mathrm{\theta }}\\ {\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1+{\mathrm{c}}_2{\mathrm{e}}^{7\mathrm{\theta }}\end{array}$

Step 3: Use the method of undetermined coefficients to find a particular solution to the differential equation.

According to the method of undetermined coefficients, assume the particular solution of equation (1),

 

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=\mathrm{\theta }({\mathrm{A}}_2{\mathrm{\theta }}^2+{\mathrm{A}}_1\mathrm{\theta }+{\mathrm{A}}_0){\mathrm{e}}^{\left(0\right)\mathrm{\theta }}\\ {\mathrm{y}}_{\mathrm{p}}={\mathrm{A}}_2{\mathrm{\theta }}^3+{\mathrm{A}}_1{\mathrm{\theta }}^2+{\mathrm{A}}_0\mathrm{\theta } ......\left(2\right)\end{array}$

Now find the derivative of the above equation,

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=3{\mathrm{A}}_2{\mathrm{\theta }}^2+2{\mathrm{A}}_1\mathrm{\theta }+{\mathrm{A}}_0\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=6{\mathrm{A}}_2\mathrm{\theta }+2{\mathrm{A}}_1\end{array}$

 

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

 

We get,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{\theta }\right)-7{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{\theta }\right)={\mathrm{\theta }}^2\\ 6{\mathrm{A}}_2\mathrm{\theta }+2{\mathrm{A}}_1-7(3{\mathrm{A}}_2{\mathrm{\theta }}^2+2{\mathrm{A}}_1\mathrm{\theta }+{\mathrm{A}}_0)={\mathrm{\theta }}^2\\ 6{\mathrm{A}}_2\mathrm{\theta }+2{\mathrm{A}}_1-21{\mathrm{A}}_2{\mathrm{\theta }}^2-14{\mathrm{A}}_1\mathrm{\theta }-7{\mathrm{A}}_0={\mathrm{\theta }}^2\\ -21{\mathrm{A}}_2{\mathrm{\theta }}^2+(-14{\mathrm{A}}_1+6{\mathrm{A}}_2)\mathrm{\theta }+(-7{\mathrm{A}}_0+2{\mathrm{A}}_1)={\mathrm{\theta }}^2\end{array}$

Step 4: Final conclusion.

Comparing all coefficients of the above equation;

 $\begin{array}{c}-21{\mathrm{A}}_2=1\Rightarrow {\mathrm{A}}_2=\frac{-1}{21}\\ -14{\mathrm{A}}_1+6{\mathrm{A}}_2=0.......\left(3\right)\\ -7{\mathrm{A}}_0+2{\mathrm{A}}_1=0.......\left(4\right)\end{array}$

 

Substitute the value of ${\mathrm{A}}_2$ in the equation (3),

 

$\begin{array}{c}-14{\mathrm{A}}_1+6\left(\frac{-1}{21}\right)=0\\ -14{\mathrm{A}}_1=\frac{2}{7}\\ {\mathrm{A}}_1=\frac{-1}{49}\end{array}$

 

Substitute the value of  $A_1$in the equation (4),

$\begin{array}{c}-7{\mathrm{A}}_0+2\left(\frac{-1}{49}\right)=0\\ {\mathrm{A}}_0=-\frac{2}{343}\end{array}$

Substitute the value of  ${\mathbf{A}}_0\mathbf{,}{\mathbf{A}}_1$ and ${\mathbf{A}}_2$ in the equation (2),

 

$\begin{array}{c}{\mathbf{y}}_p\mathbf{=}{\mathbf{A}}_2{\mathbf{\theta }}^3\mathbf{+}{\mathbf{A}}_1{\mathbf{\theta }}^2\mathbf{+}{\mathbf{A}}_0\mathbf{\theta }\\ {\mathbf{y}}_p\mathbf{=}\mathbf{-}\frac{1}{21}{\mathbf{\theta }}^3\mathbf{-}\frac{1}{49}{\mathbf{\theta }}^2\mathbf{-}\frac{2}{343}\mathbf{\theta }\end{array}$

 

Therefore, the particular solution of equation (1),

 

${\mathbf{y}}_p\mathbf{=}\mathbf{-}\frac{1}{21}{\mathbf{\theta }}^3\mathbf{-}\frac{1}{49}{\mathbf{\theta }}^2\mathbf{-}\frac{2}{343}\mathbf{\theta }$