Question

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

$2y\text{'}\text{'}\text{'}-y\text{'}\text{'}-10y\text{'}-7y=0$

Short answer

The general solution for the differential equation with x as the independent variableis .$y=C_1{e}^{-x}+C_2{e}^{\frac{3+\sqrt{65}}{4}x}+C_3{e}^{\frac{3-\sqrt{65}}{4}x}$

Step-by-step solution

Auxiliary equation:

The given differential equationis $2y\text{'}\text{'}\text{'}-y\text{'}\text{'}-10y\text{'}-7y=0$. To solve this equation, we look at its auxiliary equation which is $2m^3-m^2-10m-7=0$. Observe that -1 is a solution of this equation. So, $2m^3-m^2-10m-7=(m+1)(2m^2-3m-7)$

Inspecting the sum further:

To get the other two roots of auxillary equation, we need to solve$2m^2-3m-7=(m+1)(2m^2-3m-7)$ . We have,

$m=\frac{3\pm \sqrt{9+56}}{4}=\frac{3\pm \sqrt{65}}{4}$

General solution:

We have m = -1,$\frac{3\pm \sqrt{65}}{4}$. From (7) of 328 and (18) of page 330, we conclude that the general solution of the given differential equation is $y=C_1{e}^{-x}+C_2{e}^{\frac{3+\sqrt{65}}{4}x}+C_3{e}^{\frac{3-\sqrt{65}}{4}x}$where $C_1,C_2,C_3$ are arbitrary constants.

The solution of the given differential equation is $y=C_1{e}^{-x}+C_2{e}^{\frac{3+\sqrt{65}}{4}x}+C_3{e}^{\frac{3-\sqrt{65}}{4}x}$ , where $C_1,C_2,C_3$are arbitrary constant.

Hence the final solution is $y=C_1{e}^{-x}+C_2{e}^{\frac{3+\sqrt{65}}{4}x}+C_3{e}^{\frac{3-\sqrt{65}}{4}x}$