Finding the general solution of differential function y "- 2Y = e ^ x That's the process of finding the derivative of Y

Finding the general solution of differential function y "- 2Y = e ^ x That's the process of finding the derivative of Y

Characteristic equation R ^ 2-2 = 0, r = ± √ 2
The general solution of homogeneous equation is y = C1E ^ (√ 2x) + c2e ^ (- √ 2x)
The special solution is y * = 3E ^ X
So the general solution is y = C1E ^ (√ 2x) + c2e ^ (- √ 2x) + 3E ^ X

Seek the general solution of differential x ^ 2Y '= (x-1) y, and seek the great God in the exam

x^2 dy/dx=(x-1)y
=> dy/y=(x-1)/x^2dx
Integral on both sides at the same time

General solution of total differential Y "- 6y '- 9y = 0

Characteristic equation: λ ^ 2-6, λ - 9 = 0
(λ-3)^2=18
λ=3±3√2
So the general explanation is as follows:
y=C1*e^((3+3√2)x)+C2*e^((3-3√2)x)