Q40E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find three linearly independent solutions (see Problem 35) of the given third-order differential equation and write a general solution as an arbitrary linear combination of it. $\mathit{y}\text{'}\text{'}\text{'}\mathbf{-}\mathbf{7}\mathit{y}\text{'}\text{'}\mathbf{+}\mathbf{7}\mathit{y}\text{'}\mathbf{+}\mathbf{15}\mathit{y}\mathbf{=}\mathbf{0}$.

Step-by-Step Solution

Answer

$y\left(t\right)=c_1{e}^{-t}+c_2{e}^{5t}+c_3{e}^{3t}$ is the solution of the given equation $y\text{'}\text{'}\text{'}-7y\text{'}\text{'}+7y\text{'}+15y=0$.

Step 1: Differentiate the value of y

Given differential equation is $y\text{'}\text{'}\text{'}-7y\text{'}\text{'}+7y\text{'}+15y=0$

Let $y=e^{rt}$, then $y\text{'}\left(t\right)=r{e}^{rt}$ and

$$\begin{gathered} y\text{'}\text{'}\left(t\right)=r^2{e}^{rt} \\ y\text{'}\text{'}\text{'}=r^3{e}^{rt} \end{gathered}$$

 

Then the auxiliary equation is:

$$\begin{gathered} r^3{e}^{rt}-7r^2{e}^{rt}+7r{e}^{rt}+15e^{rt}=0 \\ \left(r^3-7r^2+7r+15\right)e^{rt}=0 \\ r^3-7r^2+7r+15=0 \end{gathered}$$ 

Step 2: Substitute the value for r

Let substitute $r=1$

$1^3-7(1{)}^2+7\left(1\right)+15\ne 0$ 

 

Now substitute $r=-1$        

${(-1)}^3-7(-1{)}^2+7(-1)+15=0$ 

 

Therefore, $r=-1$ is a solution to the auxiliary equation:

 $$\begin{gathered} r^3-7r^2+7r+15=(r+1)\left(r^2-8r+15\right) \\ (r+1)\left(r^2-8r+15\right)=0 \end{gathered}$$

 

Therefore, the solutions are;

$r=-1$ 

 

And 

$$\begin{gathered} \left(r-5\right)\left(r-3\right)=0 \\ r=5,3 \end{gathered}$$ 

 

Thus, the solution is:

 $y\left(t\right)=c_1{e}^{-t}+c_2{e}^{5t}+c_3{e}^{3t}$