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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
If $y = P e^{ax} + Qe^{bx},$ show that $\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} - (\text{a + b })\frac{\text{dy}}{\text{dx}} + \text{aby} = 0 .$
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Answer

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

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