Question

Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b: $y = a e{3x} + b e{- 2x}$

Answer 1

Answer:

The reuired differential equation is

$\frac{d^2y}{dx^2}-\frac{dy}{dx}-6y=0$

Step-by-step explanation:

Concept:

Differential equations are formed by eliminating arbitrary constants occur in the equation.

$y=a.e^{3x}+b.e^{-2x}\\\\y=e^{-2x}[a.e^{5x}+b]\\\\e^{2x}y=a.e^{5x}+b$

Differentiate with respect to x

$[tex]e^{2x}\frac{dy}{dx}+y.2e^{2x}=5ae^{5x}\\\\e^{2x}[\frac{dy}{dx}+2y]=5ae^{5x}\\\\\frac{e^{2x}}{e^{5x}}[\frac{dy}{dx}+2y]=5a\\\\e^{-3x}[\frac{dy}{dx}+2y]=5a\\\\Differentiate\:with\:respect \:to \:x\\\\e^{-3x}[\frac{d^2y}{dx^2}+2\frac{dy}{dx}]+(-3)e^{-3x}[\frac{dy}{dx}+2y]=0\\\\e^{-3x}[\frac{d^2y}{dx^2}+2\frac{dy}{dx}--3\frac{dy}{dx}-6y]=0\\\\e^{-3x}[\frac{d^2y}{dx^2}-\frac{dy}{dx}-6y]=0$[/tex]

since\:$\:e^{-3x}\neq0,$

we have

$\frac{d^2y}{dx^2}-\frac{dy}{dx}-6y=0$