Variable Coefficient Equations

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 4.7 – Variable Coefficient Equations

Expected Educational Results

• Objective 14–1: I can identify variable-coefficient equations.
• Objective 14–2: I can find a solution to a homogeneous Cauchy-Euler equation.
• Objective 14–3: I can find a solution to a nonhomogeneous Cauchy-Euler equation.
• Objective 14–4: Given a solution, I can apply the Reduction of Order Theorem to find another solution to a Cauchy-Euler equation.
• Objective 14-5: I can find a solution to a Cauchy-Euler IVP.

Variable Coefficient Equations

Method of Solving Homogeneous Cauchy-Euler Equations

Let $\displaystyle at^2y^{\,\prime\prime}(t)+bty^{\,\prime}(t)+cy(t)=0$.

1. From Activity 02, the characteristic equation is $ar^2+(b-a)r+c=0$.

2. Find the solutions to the characteristic equation, $r_1$ and $r_2$.

   • If $r_1\ne r_2$, then the linearly independent solutions are $y_1(t)=t^{r_1}$ and $y_2(t)=t^{r_2}$ and the homogeneous solution is $y_h(t)=c_1 t^{r_1}+c_2 t^{r_2}$
   • If $r_1=r_2$, then the linearly independent solutions are $y_1(t)=t^{r}$ and $y_2(t)=t^{r}\ln{(t)}$ and the homogeneous solution is $y_h(t)=c_1 t^{r_1}+c_2 t^{r}\ln{(t)}$
   • If $r_1$ and $r_2$ are complex numbers in the form $\alpha\pm i\beta$, then the linearly independent solutions are $y_1(t)=t^{\alpha}\cos{(\beta \ln{(t)})}$ and $y_2(t)=t^{\alpha}\sin{(\beta \ln{(t)})}$ and the homogeneous solution is $y_h(t)=c_1 t^{\alpha}\cos{\left(\beta \ln{(t)}\right)}+c_2 t^{\alpha}\sin{\left(\beta \ln{(t)}\right)}$

Investigation 03

1. For what values of $t$ is $y(t)=t^{r}$ not a solution to $\displaystyle at^2y^{\,\prime\prime}(t)+bty^{\,\prime}(t)+cy(t)=0$? Explain.
2. For what values of $r$ is $y(t)=t^{r}$ a solution to $\displaystyle at^2y^{\,\prime\prime}(t)+bty^{\,\prime}(t)+cy(t)=0$? Explain.
3. Suppose $y(t)=t^{r}$ is a solution to $\displaystyle at^2y^{\,\prime\prime}(t)+bty^{\,\prime}(t)+cy(t)=0$. Find the associated characteristic equation.

Example 01: Solve $\displaystyle t^2 y^{\,\prime\prime}-ty^{\,\prime}+3y=0$, $t>0$

Solution:

Note that the coefficients of each $y$-term is a polynomial function of the independent variable in which the degree of the polynomial term is equivalent to the degree of the derivative of $y$. Thus, this ode is a Cauchy-Euler equation.

Identify the coefficients from the ode: $a=1$, $b=-1$, $c=3$. and $b-a=(-1)-1=-2$.

The, characteristic equation for the above Cauchy-Euler equation is: $r^2-2r+3=0$. Using Mathematica, the roots of the characteristic equation are: $r=1 \pm i\sqrt{2}$, where $\alpha=1$ and $\beta=\sqrt{2}$.

Thus, the homogeneous solution is $\displaystyle y_h(t)=c_1 t^{1}\cos{(\sqrt{2} \ln{(t)})}+c_2 t^{1}\sin{(\sqrt{2} \ln{(t)})}$.

Investigation 04

1. Use the Wronskian to verify $y_1(t)=t^{r_1}$ and $y_2(t)=t^{r_2}$, where $r_1\ne r_2$, are linearly independent. Explain.
2. Why is the condition, $r_1\ne r_2$, above necessary? Explain.
3. Use the Wronskian to verify $y_1(t)=t^{r}$ and $y_2(t)=t^{r}\ln{(t)}$ are linearly independent. Explain.

Investigation 05

Solve the following homogeneous DEs:

1. $\displaystyle t^2 y^{\,\prime\prime}+7ty^{\,\prime}-7y=0$, $t>0$.
2. $\displaystyle t^2 y^{\,\prime\prime}-3ty^{\,\prime}+4y=0$, $t>0$.
3. $\displaystyle t^2 y^{\,\prime\prime}+4ty^{\,\prime}+4y=0$, $t>0$.
4. $\displaystyle y^{\,\prime\prime}-\frac{1}{t}y^{\,\prime}+\frac{5}{t^2}y=0$, $t>0$.
5. $\displaystyle (t-2)^2 y^{\,\prime\prime}-7(t-2) y^{\,\prime}+7y=0$, $t>2$. Hint: Let $u=t-2$.

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Last Modified: Wednesday, 14 October 2020 11:42 EDT