Question

If $y=c_1{e}^{2x}+c_2{e}^x+c_3{e}^{-x}$ satisfies the differential equation $\frac{d^{3}y}{{dx}^3}+a\frac{d^{2}y}{d{x}^2}+b\frac{dy}{dx}+cy=0$, then $\frac{a^3+b^3+c^3}{abc}$ is equal to:

• A. $\frac{1}{4}$
• B. $-\frac{1}{4}$
• C. $\frac{1}{2}$
• D. $-\frac{1}{2}$

Answer 1

Answer: (B)

$-\frac{1}{4}$

Given equation is $y=c_1{e}^{2x}+c_2{e}^x+c_3{e}^{-x}$

Now,

differentiating with respect to $x$ gives,

$\frac{dy}{dx}=2c_1{e}^{2x}+c_2{e}^x-c_3{e}^{-x}$

differentiating with respect to $x$ again gives,

$\frac{d^{2}y}{d{x}^2}=4c_1{e}^{2x}+c_2{e}^x+c_3{e}^{-x}$

differentiating with respect to $x$ again gives,

$\frac{d^{3}y}{d{x}^3}=8c_1{e}^{2x}+c_2{e}^x-c_3{e}^{-x}$

Substituting in the main equation gives,

$e^{2x}(8c_1+4a{c}_1+2b{c}_1+c{c}_1)+e^x(c_2+a{c}_2+b{c}_2+c{c}_2)+e^{-x}(-c_3+a{c}_3-b{c}_3+c{c}_3)=0$

As,powers of $e$ are non zero,the coefficients should be zero in order to make the sum zero.

$\therefore$equating each coeffcient to $0$ and solving them gives,

$a=-2,b=-1,c=2$

Substituting in the required result gives,

$\frac{a^3+b^3+c^3}{abc}=-\frac{1}{4}$