Answer to Question #331388 in Differential Equations for Prem

Question #331388

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


1
Expert's answer
2022-04-21T13:59:38-0400

"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}\n -5C=-6\\\\\n 3C-5D=3\\\\\n 6C+2D-5E=6\\\\\n 2D+E-5F=0\n\\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}".


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