Solving a special type of a second order differential equation

[Jim4592]

Homework Statement

y''[x] = y'[x] + x

Homework Equations

We were taught two special types of second order diff. equations:

Type 1: Supposed to be when x is missing
v = y'[x]
v'[x] = y''[x]

Type 2: Supposed to be when y is missing
v = y'[x]
v v'[x] = y''[x]

The Attempt at a Solution

The answer key reads:

v'[x] - v = x

d/dx{e^-x v} = x e^-x

e^-x v = C1 * e^-x -x -1

v = C1 * e^x -x -1

y[x] = c2 - c1 * e^-x - (.5)x^2 - x


I don't understand why we used type 1 to solve this problem since x is clearly stated in the problem. I was hoping someone could explain, thanks.

 

[LCKurtz]

Are you sure you have listed the two types correctly? Usually, if y is missing you would do:

v = y'
v' = y''

so your second order equation in y becomes a first order equation in v. That is the type of equation you have and also the method your manual gives.