Question

If $y'' - 3y' + 2y = 0$ where $y(0) = 1$, $y'(0) = 0$, then the value of $y$ at $x \,= log_e \,2$ is

• A. 1
• B. -1
• C. 2
• D. 0

Answer 1

Answer: (D)

$\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y=0$
The corresponding equation is $m^{2}-3 m+2=0$
$\therefore $ General solution of given equation
$y=A e^{x}+B e^{2 x}$
$y^{'}=A e^{x}+2 B e^{2 x}$
At $x=0, y=1 $
$\Rightarrow A+B=1$
and $x=0, y^{'}=0$
$ \Rightarrow A+2 B=0$
Solving these equation $A=2,\, B=1$
$\therefore y=2 e^{x}-e^{2 x}$
At $x=\log 2, \,y=0$