Question

If $y\left(x\right)=\lambda e^{2x}+e^{\alpha x},\alpha \ne 2$ is a solution of the differential equation
$\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-6=0$
Satisfying $\frac{dy}{dx}\left(0\right)=5$ then y(0) is equal to

• A. 1
• B. 4
• C. 5
• D. 9

Answer 1

The correct option is C 5

$\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}-6=0$

Solving this differential equation,

$D^2+D-6=0$

$\Rightarrow D^2+3D-2D-6$

$\Rightarrow (D+3)(D-2)=0,D=-3,D=2$

General solution

$y\left(x\right)=C_1{e}^{2x}+C_2{e}^{-3x}$

Comparing with the solution

$y\left(x\right)=\lambda e^{2x}+e^{\alpha x}=C_1{e}^{2x}+C_2{e}^{-3x}$

$\lambda =C_1,C_2=1and\alpha =-3$

So, $y\left(x\right)=\lambda e^{2x}+e^{-3x}$

$\frac{dy}{dx}=\lambda e^{2x}\cdot 2+e^{-3x}\cdot (-3)$
$\because {\begin{array}{c}\frac{dy}{dx}\end{array}|}_{x=0}=5$

$\Rightarrow 5=2\lambda e^0-3e^0$

$\Rightarrow 2\lambda =8,\lambda =4$

we get,

$y\left(x\right)=4e^{2x}+e^{-3x}$

At x = 0
$y\left(0\right)=4e^0+e^0$

y(0) = 5