Question 4.8.1: A Second-Order Linear Differential Equation Show that both y...

For the function y_1 = e^x, you have y_{1}^{′} = e^x and y_1″ = e^x. So,

y_1″ – y_1= e^x – e^x = 0
which means that y_1 = e^x is a solution of the differential equation. Similarly, for y_2 = e^{-x}, you have

y_{2}^{′} = e^x and y_2″ = e^x.
This implies that

y_2″ – y_2= e^x – e^x = 0.
So, y_2 = e^{-x} is also a solution of the linear differential equation.