Question

The differential equation $\frac{d^{2}y}{d{x}^2}+\frac{6dy}{dx}+9y=6e^{-3x}$ has boundary conditions $y\left(0\right)=0,y(+1)=6e^{-3}$

then $y(-1)$ is

• A. 0

Answer 1

The correct option is A 0

Given ,

$[D^2+6D+9]y=6e^{-3x}$

So, auxiliary equation is,

$\Rightarrow m^2+6m+9=0$

$(m+3{)}^2=0$

$\Rightarrow m=-3,-3$

So, complimentary function, C.F is $\to [C_1+C_{2}x]e^{-3x}$ ... (i)

Now, $P.I=\frac{1}{D^2+6D+9}6e^{-3x}$

For D = -3 case fails so ,

$P.I=\frac{6x{e}^{-3x}}{2D+6}$

Again for D = -3 case fails so,

$P.I=\frac{6x^2{e}^{-3x}}{2}=3x^2{e}^{-3x}$ ... (ii)

Using (1) and (2)

So, complete solution is ,

$y=[C_1+C_{2}x]e^{-3x}+3x^2{e}^{-3x}$

$\Rightarrow y=[C_1+C_{2}x+3x^2]e^{-3x}$

Now, $y\left(0\right)=0$

$\Rightarrow 0=\left[C_1\right]e^{-3.0}$

$\Rightarrow C_1=0$

Also, $y\left(1\right)=6e^{-3}$

$\Rightarrow [C_2+3]e^{-3}=6e^{-3}$

$\Rightarrow C_2=3$

So, $y=[3x+3x^2]e^{-3x}$

$\Rightarrow y(-1)=[-3+3]e^3=0$