# Question in undetermined coefficient method for ODE

How should I formulate particular solution of this ODE? I want to use method of undetermined coefficients.

$$y'' - y = e^x \\ y_H = C_1 e^x +C_2 x e^x$$ $y_H, y_P$ are homogeneous and particular solutions respectively.

From method of undetermined coefficients, I assume $y_P = Ax^2 e^x + Bxe^x$ Is it valid to include first power of x? It is part of homogeneous solution. $$y^{'}_P=Be^x + ax^2 e^x + (2A+B)xe^x \\ y^{''}_P = e^x(2A+2B) + xe^x(4A+B) + Ax^2 e^x$$ Substituting above values in given ODE, $$y^{''}_P - y_P = e^x (2A+2B) + xe^x (4A) = e^x$$ Comparing coefficients, $$A=0 \\ B=\frac{1}{2} \\ \therefore y_P =\frac{1}{2} xe^x$$

This result is wrong as it is homogeneous solution. Now, I take another particular solution.

$$y_P = Ax^3 e^x + Bx^2 e^x \\ y^{'}_P= x^2 e^x (3A + B) + 2Bxe^x + Ax^3 e^x \\ y^{''}_P= 2B e^x + x e^x (6A + 4B) + x^2 e^x (6A + B) + A e^x x^3 \\ \therefore y^{''}_P - y_P = 2B e^x + x e^x (6A + 4B) + 6A e^x x^2 = e^x \\$$ Comparing coefficients, $$A=0 \\ 6A + 4B = 0 \\ B=\frac{1}{2}\\$$ Now, these conditions seem inconsistent. Can someone please explain?

• I got the mistake, thanks for replies. I went wrong in finding homogeneous solution. – gyeox29ns Oct 23 '14 at 1:05

Here, there is already a problem with the characteristic equation: $$r^2 - 1 = 0 \iff r = \pm 1$$

I recommend you to take a look at the variation of constants:

$$y_P = Ae^x + Be^{-x}$$with $A,B$ such as $$A'e^x + B'e^{-x} = 0$$

When you write the derivatives you get: $$y_P' = Ae^x - Be^{-x} \\ y_P'' = Ae^x + A'e^x + Be^{-x} - B'e^{-x}$$ so $$A'e^x + B'e^{-x} = 0\\ A'e^x - B'e^{-x} = e^x \\\implies A' = \frac 12, B' = -\frac 12 e^{2x}; \\ A = \frac 12x + a, B = -\frac 14 e^{2x} + b$$ and finally $$y_P = Ae^x + Be^{-x} = \frac 12xe^x -\frac 14 e^{x} + ae^x + be^{-x}, (a,b)\in\Bbb R^2$$

or after simplification: $$y_P = Ae^x + Be^{-x} = \frac 12xe^x + a'e^x + be^{-x}, (a',b)\in\Bbb R^2$$

$y_p=C_1e^{x}+C_2e^{-x}$ and for the special solution of the inhomogeneous equation make the ansatz $(Ax+B)e^{x}$

• the part $Be^x$ is useless. – mookid Oct 22 '14 at 22:25