Q49E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The reduction of order procedure can be used more generally to reduce a homogeneous linear $(\mathbf{n}\mathbf{-}\mathbf{1})$  th-order equation to a homogeneous linear  $(\mathbf{n}\mathbf{-}\mathbf{1})$th-order equation. For the equation, $\mathbf{ty}\text{'}\text{'}\text{'}\mathbf{-}\mathbf{ty}\text{'}\text{'}\mathbf{+}\mathbf{y}\text{'}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$ which has $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^t$ as a solution, use the substitution $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{v}\left(\mathbf{t}\right)\mathbf{f}\left(\mathbf{t}\right)$ to reduce this third-order equation to a homogeneous linear second-order equation in the variable $\mathbf{w}\mathbf{=}\mathbf{v}\text{'}$.

 

Step-by-Step Solution

Answer

The second-order homogeneous equation is $\mathbf{tw}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^t\mathbf{+}\mathbf{2}{\mathbf{e}}^t\mathbf{w}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{e}}^t\mathbf{w}\mathbf{=}\mathbf{0}$.

 

Step 1: Differentiate the values of y

Given differential equation is $\mathrm{ty}\text{'}\text{'}\text{'}-\mathrm{ty}\text{'}\text{'}+\mathrm{y}\text{'}-\mathrm{y}=0$ and  $\mathrm{f}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}.$

 

Now,

$\begin{array}{c}\mathrm{y}\left(\mathrm{t}\right)=\mathrm{v}\left(\mathrm{t}\right)\mathrm{f}\left(\mathrm{t}\right)\\ ={\mathrm{ve}}^{\mathrm{t}}\end{array}$ 

 

 

Find the derivatives of the above result.

 $\begin{array}{c}\mathrm{y}\text{'}\left(\mathrm{t}\right)=\frac{\mathrm{d}}{\mathrm{dt}}\left({\mathrm{ve}}^{\mathrm{t}}\right)\\ =\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\\ \mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\right)\\ =\mathrm{v}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\\ =\mathrm{v}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+2\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\end{array}$

 

Now find the third derivative of $\mathrm{y}$.

 $\begin{array}{c}\mathrm{y}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)=\mathrm{v}\text{'}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+2\mathrm{v}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+2\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\\ =\mathrm{v}\text{'}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+3\left(\mathrm{v}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\right)+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\end{array}$

Step 2: Substitute the values.

Substitute the above values in the differential equation;

$\begin{array}{c}\mathrm{ty}\text{'}\text{'}\text{'}-\mathrm{ty}\text{'}\text{'}+\mathrm{y}\text{'}-\mathrm{y}=\mathrm{tv}\text{'}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+3\left(\mathrm{v}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\right)+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}-\mathrm{t}\left(\mathrm{v}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+2\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\right)\\ +\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}-\left({\mathrm{ve}}^{\mathrm{t}}\right)\\ =\mathrm{tv}\text{'}\text{'}\text{'}{\mathrm{e}}^{\mathrm{t}}+2{\mathrm{e}}^{\mathrm{t}}\mathrm{v}\text{'}\text{'}+(\mathrm{t}+1){\mathrm{e}}^{\mathrm{t}}\mathrm{v}\text{'}\end{array}$

Now use $\mathrm{w}=\mathrm{v}\text{'}$in the above result;

 

$\mathrm{tw}\text{'}\text{'}{\mathrm{e}}^{\mathrm{t}}+2{\mathrm{e}}^{\mathrm{t}}\mathrm{w}\text{'}+(\mathrm{t}+1){\mathrm{e}}^{\mathrm{t}}\mathrm{w}=0$

 

Therefore, the second order homogeneous equation is $\mathbf{tw}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^t\mathbf{+}\mathbf{2}{\mathbf{e}}^t\mathbf{w}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{e}}^t\mathbf{w}\mathbf{=}\mathbf{0}$.