Question

Solve the following using the method of undetermined coefficients. (d ^ 2 * y)/(d * t ^ 2) + (dy)/(dt) - 5y = - 6x ^ 3 + 3x ^ 2 + 6x

Answer 1

$y''+y'-5y=-6x^3+3x^2+6x$ (1)

The solutions to a nonhomogeneous equation are of the form

$y(x)= y_h(x) + y_p(x)$ ,

where $y_h$ is the general solution to the associated homogeneous equation and $y_p$ is a particular solution.

The associated homogeneous equation:

$y''+y'-5y=0$

The general solution of this equation is determined by the roots of the characteristic equation:

$\lambda^2+\lambda-5=0$

$D=1^2-4\cdot1\cdot(-5)=21$

$\lambda=\frac{-1\pm\sqrt{21}}{2}$

$y_h(x)=Ae^{\frac{-1+\sqrt{21}}{2}x}+Be^{\frac{-1-\sqrt{21}}{2}x}$ , where $A$ and $B$ are constants.

The particular solution of the differential equation:

$y_p(x)=Cx^3+Dx^2+Ex+F$

$y_p'(x)=3Cx^2+2Dx+E$

$y_p''(x)=6Cx+2D$

Substitution $y_p$ , $y_p'$ , and $y_p''$ into (1) gives:

$6Cx+2D+3Cx^2+2Dx+E-$$5(Cx^3+Dx^2+Ex+F)=-6x^3+3x^2+6x$

The last equation must be valid for all values of x, so the coefficients with the like powers of x

 in the right and left sides must be identical:

$\begin{cases} -5C=-6\\ 3C-5D=3\\ 6C+2D-5E=6\\ 2D+E-5F=0 \end{cases}$

We find from this system that 

 $C=\frac65$ ; $D=\frac{3}{25}$ ; $E=\frac{36}{125}$ ; $F=\frac{66}{625}$

 As a result, the particular solution is written as

$y_p(x)=\frac65x^3+\frac{3}{25}x^2+\frac{36}{125}x+\frac{66}{625}$

Answer: $y(x)=Ae^{\frac{-1+\sqrt{21}}{2}x}+Be^{\frac{-1-\sqrt{21}}{2}x}+\frac65x^3+$$\frac{3}{25}x^2+\frac{36}{125}x+\frac{66}{625}$.