91Ó°ÊÓ

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, $2m^3−m^2−10m−7=(m+1)(2m^2−3m−7)$

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,

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

We have m = -1,$\frac{3Â\pm \sqrt{65}}{4}$. 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 $y=C_1{e}^{−x}+C_2{e}^{\frac{3+\sqrt{65}}{4}x}+C_3{e}^{\frac{3−\sqrt{65}}{4}x}$