If $y^{'} - 3 y^{'} + 2 y = 0$ where $y (0) = 1, y^{'} (0) = 0$ , then the value of y at $x = l o g 2$ is

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Solution

The correct option is C $0$
$\frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + 2 y = 0$
The corresponding equation is $m^{2} - 3 m + 2 = 0$
$∴$ 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 $A = 2, B = 1$
$∴ y = 2 e^{x} - e^{2 x}$
At $x = l o g 2, y = 0$

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