Q50E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

The equation $\mathbf{ty}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{t}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$has $\mathbf{f}\left(\mathbf{t}\right)\mathbf{=}\mathbf{t}$ 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 $\mathbf{w}\mathbf{=}\mathbf{v}\mathbf{\text{'}}$.

Step-by-Step Solution

Answer

The second-order homogeneous equation is ${\mathrm{t}}^2\mathrm{w}\text{'}\text{'}{\mathrm{e}}^{\mathrm{t}}+(4\mathrm{t}-{\mathrm{t}}^2){\mathrm{e}}^{\mathrm{t}}\mathrm{w}\text{'}+({\mathrm{t}}^2-2\mathrm{t}+2){\mathrm{e}}^{\mathrm{t}}\mathrm{w}=0$. 

Step 1: Differentiate the value of y

Given differential equation is $\mathrm{ty}\text{'}\text{'}\text{'}+(1-\mathrm{t})\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}-\mathrm{y}=0$ 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 $y$.

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

Step 2: Substitute the values

Substitute these in the differential equation;

$\begin{array}{c}\mathrm{ty}\text{'}\text{'}\text{'}+(1-\mathrm{t})\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}-\mathrm{y}=\mathrm{t}(\mathrm{v}\text{'}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+3\left(\mathrm{v}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\right)+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+(1-\mathrm{t})\left(\mathrm{v}\text{'}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+2\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\right)\\ +\mathrm{t}\left(\mathrm{v}\text{'}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}+\mathrm{v}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{t}}\right)-\left({\mathrm{ve}}^{\mathrm{t}}\right)\\ ={\mathrm{t}}^2\mathrm{v}\text{'}\text{'}\text{'}{\mathrm{e}}^{\mathrm{t}}+(4\mathrm{t}-{\mathrm{t}}^2){\mathrm{e}}^{\mathrm{t}}\mathrm{v}\text{'}\text{'}+({\mathrm{t}}^2-2\mathrm{t}+2){\mathrm{e}}^{\mathrm{t}}\mathrm{v}\text{'}\end{array}$

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

 

${\mathrm{t}}^2\mathrm{w}\text{'}\text{'}{\mathrm{e}}^{\mathrm{t}}+(4\mathrm{t}-{\mathrm{t}}^2){\mathrm{e}}^{\mathrm{t}}\mathrm{w}\text{'}+({\mathrm{t}}^2-2\mathrm{t}+2){\mathrm{e}}^{\mathrm{t}}\mathrm{w}=0$

 

Therefore, the second-order homogeneous equation is:

 ${\mathbf{t}}^2\mathbf{w}\mathbf{\text{'}}\mathbf{\text{'}}{\mathbf{e}}^t\mathbf{+}(\mathbf{4}\mathbf{t}\mathbf{-}{\mathbf{t}}^2){\mathbf{e}}^t\mathbf{w}\mathbf{\text{'}}\mathbf{+}({\mathbf{t}}^2\mathbf{-}\mathbf{2}\mathbf{t}\mathbf{+}\mathbf{2}){\mathbf{e}}^t\mathbf{w}\mathbf{=}\mathbf{0}$.

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