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

$y = P e^{ax} + Q e^{bx}$^$\Rightarrow\frac{\text{dy}}{\text{dx}}$
$= a P e^{ax} + b Q e^{bx}$^
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=$
$a2 p e_{ax} + b2 Q$ ebx
$\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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