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