Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution for the differential equation with x as the independent variable:

 $y\text{'}\text{'}\text{'}-y\text{'}\text{'}+2y=0$

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 given differential equationis  $y\text{'}\text{'}\text{'}-y\text{'}\text{'}+2y=0$. To solve this equation, we look at its auxiliary equation which is $m^3-m^2+2=0$  . Observe that -1 is a solution of this equation. So,

 $m^3-m^2+2=(m+1)(m^2-2m+2)$

Step 2: Inspecting the sum further:

To get the other two roots of auxillary equation, we need to solve  $m^3-m^2+2=0$ . We have,

 $m=\frac{2\pm \sqrt{4-8}}{2}=1\pm i$

Step 3: General solution:

We have m = -1,$1\pm i$ .From (7) of 328 and (18) of page 330, we conclude that the general solution of the given differential equation is $y=C_1{e}^{-x}+C_2{e}^x\cos x+C_3{e}^x\sin x$  where  $C_1,C_2,C_3$ are arbitrary constants.

The solution of the given differential equation is$y=C_1{e}^{-x}+C_2{e}^x\cos x+C_3{e}^x\sin x$  , where  $C_1,C_2,C_3$ are arbitrary constant.

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