Question

Using the method of undermined coefficients, find the general solution of the differential equation y^iv-2y'''+2y''=3e^-x+2e^-xx+e^-x sin x

Answer 1

$$y^{\mathrm{iv}}-2y'''+2y''=3e^{-x}+2e^{-x}x+e^{-x}\sin x$$

To determine the general solution needed find a complementary and partial solutions. First, determine a partial solution.

Rewrite the right part of the equation as

$$f(x) = (3+2x)e^{-x} + e^{-x}\sin x$$

Therefore, a partial solution of the equation have a form

$$y_p(x) = (A+Bx)e^{-x} + e^{-x}(C\sin x + D\cos x)$$

Find the derivatives.

$$y'_p = Be^{-x} - (A+Bx)e^{-x} - e^{-x}((C+D)\sin x - (C-D)\cos x) \\ y''_p = -2Be^{-x} + (A+Bx)e^{-x} + e^{-x}(2D\sin x - 2C\cos x) \\ y'''_p = 3Be^{-x} - (A+Bx)e^{-x} + e^{-x}((2C-2D)\sin x + (2C+2D)\cos x) \\ y^{\text{iv}}_p = -4Be^{-x} + (A+Bx)e^{-x} - e^{-x}(4C\sin x +4D\cos x)$$

Substituting these into the equation, obtain

$$-14Be^{-x} + 5(A+Bx)e^{-x} + e^{-x}((8D-8C)\sin x - (8D + 8C)\cos x) \\= 3e^{-x}+2xe^{-x}+e^{x}\sin x$$

Rewrite the left part.

$$(5A-14B)e^{-x} + 5Bxe^{-x} + e^{-x}((8D-8C)\sin x - (8D + 8C)\cos x) \\= 3e^{-x}+2xe^{-x}+e^{x}\sin x$$

Therefore,

$$5A-14B=3\\ 5B=2\\ 8D-8C=1\\ -8D-8C=0$$

Solving these linear system, get that $A= \frac{43}{25}, B = \frac{2}{5}, C = -\frac{1}{16}, \text{ and } D = \frac{1}{16}.$

Thus, the partial solution is

$$y_p(x)=\left(\frac{43}{25} + \frac{2}{5}x \right)e^{-x} + e^{-x}\left( \frac{1}{16}\cos x - \frac{1}{16}\sin x \right)$$

Now, determine a complementary solution. Solve the characteristic equation.

$$\lambda^4 - 2\lambda^3 + 2\lambda^2=0\\ \lambda^2(\lambda^2 - 2\lambda + 2) =0$$

This equation have the roots $\lambda = 0 \text{(multiplicity 2) and } \lambda = 1 \pm i.$ Therefore, the complementary solution is

$$y_c(x) = C_1 + C_2x + C_3e^x\cos x + C_4e^x\sin x$$

So, the genaral solution is

$$y(x) = C_1 + C_2x + C_3e^x\cos x + C_4e^x\sin x \\+ \left(\frac{43}{25} + \frac{2}{5}x \right)e^{-x} + e^{-x}\left( \frac{1}{16}\cos x - \frac{1}{16}\sin x \right)$$