Answer to Question #237423 in Differential Equations for James

Question

Eliminate the arbitrary constants.

y = x + Ae^x + Be^-2x

Answer 1

Given

$$y=x+Ae^x + Be^{-2x} \quad \cdots (i)$$

From eqn (i) above,

$$y-x=Ae^x + Be^{-2x} \quad \cdots (ii)$$

Since we have two arbitrary constants, we'll differentiate eqn (i) twice.

$$y'=1+Ae^x -2Be^{-2x} \quad \cdots (iii)\\ y''=Ae^x +4Be^{-2x} \quad \cdots (iv)$$

Adding eqn (iii) and eqn (iv), we have:

$$y''+y'=1+2Ae^x +2Be^{-2x}\\ y''+y'=1+2(Ae^x +Be^{-2x}) \quad \cdots (v)$$

Put (ii) in (v), we get

$$y''+y'=1+2(y-x)\\$$

Re-write

$$y''+y'-2y=1-2x$$