Question

If y = Pe^ax + Qe^bx, show that d^2y/dx^2 - (a + b)dy/dx + aby = 0

Answer 1

$\boxed{\underline{\textsf{Solution :}}}$

$\textsf{The given equation is}$

$\bold{y = Pe^{ax} + Qe^{bx}}$

$\textsf{Diff. both sides w. r. to x, we get}$

$\bold{\frac{dy}{dx} = \frac{d}{dx}(Pe^{ax} + Qe^{bx})}$

$\to \bold{\frac{dy}{dx} = \frac{d}{dx}(Pe^{ax}) + \frac{d}{dx}(Qe^{bx})}$

$\to \bold{\frac{dy}{dx} = P\frac{d}{dx}(e^{ax}) + Q\frac{d}{dx}(e^{bx})}$

$\to \boxed{\bold{\frac{dy}{dx} = Pae^{ax} + Qbe^{bx}}}$

$\{\textsf{since}\:\bold{\frac{d}{dx}(e^{mx}) = me^{mx}\}}$

$\textsf{Again, diff. with respect to x, we get}$

$\bold{\frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dx}(Pae^{ax} + Qbe^{bx})}$

$\to \bold{\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}(Pae^{ax}) + \frac{d}{dx}(Qbe^{bx})}$

$\to \bold{\frac{d^{2}y}{dx^{2}} = Pa\frac{d}{dx}(e^{ax}) + Qb\frac{d}{dx}(e^{bx})}$

$\to \boxed{\bold{\frac{d^{2}y}{dx^{2}} = Pa^{2}e^{ax} + Qb^{2}e^{bx}}}$

$\textsf{Now,}$

$\bold{\frac{d^{2}y}{dx^{2}} - (a+b)\frac{dy}{dx} + aby}$

$=\bold{(Pa^{2}e^{ax}+Qb^{2}e^{bx})}$

$\bold{- (a+b)(Pae^{ax}+Qbe^{bx})}$

$\bold{+ ab(Pe^{ax} + Qe^{bx})}$

$= \small{\bold{Pa^{2}e^{ax} + Qb^{2}e^{bx} -Pa^{2}e^{ax}-Qb^{2}e^{bx}}}$

$\small{\bold{-Pabe^{ax}-Qabe^{bx}+Pabe^{ax}+Pabe^{bx}}}$

$=\bold{0}$

$\to \small{\boxed{\boxed{\bold{\frac{d^{2}y}{dx^{2}} - (a+b)\frac{dy}{dx} + aby=0}}}}$

$\underline{\textsf{Hence, proved.}}$