Question

Prove that the function y = e^-5x is a solution of the differential equation d²y/dx² + dy/dx - 20y = 0.

\(\frac{d^2y}{dx^2}+\frac{dy}{dx}-20y=0\)

Answer 1

Given differential equation is

\(\frac{d^y}{dx^2}+\frac{dy}{dx}-20y=0\)

It's complementary equation is

m² + m - 20 = 0

⇒ (m + 5)(m - 4) = 0

⇒ m = -5, 4

\(\therefore\) its solution is y = C₁e^-5x + C₂e^4x

Thus, e^-5x is a solution of given differential equation.

Alternative:- Given differential equation is

\(\frac{d^2y}{dx^2}+\frac{dy}{dx}-20y=0\) 

Let y = e^-5x

^{⇒ \(\frac{dy}{dx}=-5e^{-5x} = -5y\)}

\(\therefore\) \(\frac{d^2y}{dx^2}=25e^{-5x}=25y\)

Now, \(\frac{d^2y}{dx^2}+\frac{dy}{dx} - 20y\)

 = 25y - 5y - 20y = 25y - 25y = 0

Hence, y = e^-5x is a solution of given differential equation.