Question

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

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

Short 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-by-step solution

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,

$m^3-m^2+2=(m+1)(m^2-2m+2)$

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,

$m=\frac{2\pm \sqrt{4-8}}{2}=1\pm i$

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