Q19E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Solve the given initial value problem for the Cauchy-Euler equation.

${\mathbf{t}}^2\mathbf{y}\text{'}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{4}\mathbf{ty}\text{'}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{y}\text{'}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{11}$

Step-by-Step Solution

Answer

The solution of the given initial value problem ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)-4\mathrm{ty}\text{'}\left(\mathrm{t}\right)+4\mathrm{y}\left(\mathrm{t}\right)=0;\mathrm{y}\left(1\right)=-2,\mathrm{y}\text{'}\left(1\right)=-11$is $\mathrm{y}=-3{\mathrm{t}}^4+\mathrm{t}$.

Step 1: Substitute the values.

Given differential equation is ${\mathrm{t}}^2\mathrm{y}\text{'}\text{'}\left(\mathrm{t}\right)-4\mathrm{ty}\text{'}\left(\mathrm{t}\right)+4\mathrm{y}\left(\mathrm{t}\right)=0$

 

Assume $\mathrm{y}={\mathrm{t}}^{\mathrm{r}}$, then we have;

 $\begin{array}{l}\mathrm{y}\text{'}={\mathrm{rt}}^{\mathrm{r}-1}\\ \mathrm{y}\text{'}\text{'}=\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}\end{array}$

 

Substitute these equations in the differential equation;

 $\begin{array}{c}{\mathrm{t}}^2\mathrm{r}(\mathrm{r}-1){\mathrm{t}}^{\mathrm{r}-2}-4{\mathrm{trt}}^{\mathrm{r}-1}+4{\mathrm{t}}^{\mathrm{r}}=0\\ \left(\mathrm{r}\right(\mathrm{r}-1)-4\mathrm{r}+4){\mathrm{t}}^{\mathrm{t}}=0\\ {\mathrm{r}}^2-5\mathrm{r}+4=0\end{array}$

 

 The auxiliary equation is  ${\mathrm{r}}^2-5\mathrm{r}+4=0$.

Step 2: Finding the roots of the auxiliary equation.

Find the roots of this equation:

 $\begin{array}{c}\mathrm{r}=\frac{5\pm \sqrt{5^2-4\times 4\times 1}}{2\times 1}\\ \mathrm{r}=\frac{5\pm \sqrt{25-16}}{2}\\ \mathrm{r}=\frac{5\pm \sqrt{9}}{2}\\ \mathrm{r}=\frac{5\pm 3}{2}\\ r=4,1\end{array}$

Hence, the general solution is $\mathrm{y}={\mathrm{c}}_1{\mathrm{t}}^4+{\mathrm{c}}_2{\mathrm{t}}^1.$

Step 3: Finding the values of c 1 , c 2

Using the given initial conditions;

$\begin{array}{l} \mathrm{y}\left(1\right)={\mathrm{c}}_1{\left(1\right)}^4+{\mathrm{c}}_2\left(1\right)\\ -2={\mathrm{c}}_1+{\mathrm{c}}_2\\ {\mathrm{c}}_1+{\mathrm{c}}_2=-2 \dots \left(1\right)\end{array}$

And we have $\mathrm{y}\text{'}\left(\mathrm{t}\right)=4{\mathrm{c}}_1{\mathrm{t}}^3+{\mathrm{c}}_2$ then:

$\begin{array}{l} \mathrm{y}\text{'}\left(1\right)=4{\mathrm{c}}_1{\left(1\right)}^3+{\mathrm{c}}_2\\ 4{\mathrm{c}}_1+{\mathrm{c}}_2=-11 \dots \left(2\right)\end{array}$

Subtract (2) from (1), we get:

$\begin{array}{l}3{\mathrm{c}}_1=-9\\ {\mathrm{c}}_1=-3\end{array}$

Therefore,

${\mathrm{c}}_2=1$

 

Thus, the solution is $\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{3}{\mathbf{t}}^4\mathbf{+}\mathbf{t}.$