If y = P e^ ax + Q e^ bx , show that d^2y dx^2 -(a+b) dy dx +aby=… - Vidyadip

Question

If$ y = P e^{ax}+ Q e^{bx},$ show that
$\frac{\text{d}^2\text{y}}{\text{dx}^2}-(\text{a}+\text{b})\frac{\text{dy}}{\text{dx}}+\text{aby}=0$

✓

Answer

$y = P e^{ax} + Q e^{bx }$
$\Rightarrow\frac{\text{dy}}{\text{dx}}$
$= a P e^{ax} + b Q e^{bx}$
$\frac{\text{d}^2\text{y}}{\text{dx}^2}= a^2p e^{_{ax}} + b^{2 }Q e^{bx}$
$\therefore\ \text{LHS}=$$\frac{\text{d}^2\text{y}}{\text{dx}^2}$– (a + b)$\frac{\text{dy}}{\text{dx}}$ +aby
$= a^2 P e^{ax} + b^2 Q e^{bx} – (a + b) \{a P e^{ax} + b Q e^{bx}\}+ ab \{P e^{ax} + Q e^{bx}\}$
$= P e^{ax} \{a^2 – a^2 – ab + ab\}+ Q e^{bx} \{b^2 – ab – b^2 + ab\}$
$= 0 + 0 = 0. = R.H.S.$

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