Q7E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

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

Step-by-Step Solution

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 1: 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, $2m^3-m^2-10m-7=(m+1)(2m^2-3m-7)$

Step 2: 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,

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

Step 3: 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  $y=C_1{e}^{-x}+C_2{e}^{\frac{3+\sqrt{65}}{4}x}+C_3{e}^{\frac{3-\sqrt{65}}{4}x}$