Question

Find a general solution of y" – 4y = 0.

Answer 1

Answer:

y = $C_{1}e^{2x} + C_{2}e^{-2x}$

Step-by-step explanation:

We are given the differential equation: y'' - 4y = 0

We have to find the general solution.

The auxiliary equation for the above differential equation can be written as:

m² - 4 = 0

We solve for m.

⇒m² = 4

⇒m = ±2

⇒$m_{1}$ = +2 and $m_{2}$ = -2

Thus, we have two distinct roots or we have two distinct values of m.

Thus, the general solution will be of the form:

y = $C_{1}e^{m_{1}x} + C_{2}e^{m_{2}x}$

y = $C_{1}e^{2x} + C_{2}e^{-2x}$