ODE 2y''-y'=1, y(0)=0, y'(0)=0

Question

Hello dears

It's such a crazy question, but I just confused that how can I find general solution!!?

2y''-y'=1, y(0)=0, y'(0)=0

I tried with these forms but couldn't find!

1) y=c1*c^x + c2*x*e^x

2)y=a0+a1*x+a2*x+a3*x^2+...

Answer 1

This second order differential linear equations has constant coefficients so you can reduce using the ansatz of y(x) = e^rx

2y'' - y' = 1

We see we have a homogeneous and particular solution set. Let's solve the homogeneous part first...

This would give you the following quadratic equation in r...

2r² - r = 0 --> r(2r - 1) = 0 --> r = 0 or r = 1/2

Therefore your homogeneous solution must represent the family of ...

y_H(x) = Ae^(x/2) + B

The particular solution must satisfy the solution of the first and second derivatives adding up to 1. Let's use a second ansatz of y_P(x) = Cx + D. This choice gives us zero from the second derivative and a constant from the first derivative...

2*(0) - C = 1 --> C = 1

y(x) = y_H(x) + y_P(x) = Ae^(x/2) + B + Cx + D

We can set D = 0 for simplicity...

y(0) = 0 --> 0 = A + B

y'(0) = 0 --> 0 = A/2 - C

A = -B

-B/2 - C = 0

Let C = 1 for simplicity...

B = -2, A = 2

y(x) = y_H(x) + y_P(x) = 2e^(x/2) -2 + x