Solve 2 second order differential equations?

Question

1) y'' - y = -2x^3 + x - 3

2) y'' + b^2(y) = sin(ab) b and a are real constants (b is not equal to a)

Answer 1

The method of doing these two is very similar so I'll do the first and leave you to do the second.

y'' - y = -2x^3 + x - 3

Solution will be y = yg + yp where yg is general solution including two constants of integration and yp is particular solution.

First find yg, i.e. solution to

yg'' - yg = 0

yg'' = yg

It is fairly easy to see that solutions could be yg = e^x or y = e^-x

Therefore the most general solution is y = a*e^x + b*e^-x.

You might like to check this by differentiation.

Now you find yp. The form of a particular solution is usually the same as the RHS in the original. This was a cubic in x so let's assume

yp = ax^3 + bx^2 + cx + d

yp' = 3ax^2 + 2bx + c

yp'' = 6ax + 2b

yp'' - yp = -ax^3 - bx^2 + (6a - c)x + (2b - d) = -2x^3 + x - 3

You should see fairly easily that a = 2, b = 0

these then give us that 12 - c = 1, c = 11 and d = 3

Particular solution is yp = 2x^3 + 11x + 3

Complete solution is y = a*e^x + b*e^-x + 2x^3 + 11x + 3

Try the second one in a similar manner.

yg for that is sum of two trig functions.

For yp note that RHS is constant.