If y = P eax + Q ebx, show that d^2y dx^2 -(a+b) dy dx +aby=0 - 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
^{$\frac{\text{d}^2\text{y}}{\text{dx}^2}=$} a²p e^_ax + b²Q e^bx
^{$\therefore\ \text{LHS}=$$\frac{\text{d}^2\text{y}}{\text{dx}^2}$}^{– (a + b)}^{$\frac{\text{dy}}{\text{dx}}$}^+aby
= a² P e^ax + b² Q e^bx – (a + b) {a P e^ax + b Q e^bx}+ ab {P e^ax + Q e^bx}
= P e^ax {a² – a² – ab + ab}+ Q e^bx {b² – ab – b² + ab}
= 0 + 0 = 0. = R.H.S.

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