Q4E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution for the differential equation with x as the independent variable:$y\text{'}\text{'}\text{'}-3y\text{'}\text{'}-y\text{'}+3y=0$

Step-by-Step Solution

Answer

The general solution for the differential equation with x as the independent variableis.$y\left(x\right)=c_1{e}^{-x}+c_2{e}^{-5x}+c_3{e}^{4x}$

Step 1: Auxiliary equation:

The auxiliary equation is $r^3+2r^2-19r-20=0$

Step 2: Inspecting the sum further:

By inspection, we find that r = -1 is a root and using polynomial division we get

$\begin{array}{l}{r}^3+2r^2-19r-20=(r+1)(r^2+r-20)\\ =(r+1)(r+5)(r-4)\\ =0\end{array}$

Step 3: General solution:

Now the roots of auxiliary equation are $r_1=-1,r_2=-5$ and . $r_3=4$Therefore, a general solution to given equation is 

$y\left(x\right)=c_1{e}^{-x}+c_2{e}^{-5x}+c_3{e}^{4x}$

Hence the final solution is $y\left(x\right)=c_1{e}^{-x}+c_2{e}^{-5x}+c_3{e}^{4x}$