Answer to Question #237105 in Differential Equations for Wajiha

Question #237105
Y"+4y'-y=14+6e2x
1
Expert's answer
2021-09-14T11:37:01-0400

The homogeneous differential equation


"y''+4y'-y=0"

The corresponding (auxiliary) equation


"r^2+4r-1=0"

"D=(4)^2-4(1)(-1)=20"

"r=\\dfrac{-4\\pm\\sqrt{20}}{2(1)}=-2\\pm \\sqrt{5}"

The general solution of the homogeneous differential equation is


"y=C_1e^{(-2-\\sqrt{5})x}+C_2e^{(-2+\\sqrt{5})x}"


Find the particular solution of the non-homogeneous differential equation


"y_p=A+Be^{2x}"

"y_p'=2Be^{2x}"

"y_p''=4Be^{2x}"

Substitute


"4Be^{2x}+4(2Be^{2x})-(A+Be^{2x})=14+6e^{2x}"

"-A+11Be^{2x}=14+6e^{2x}"

"A=-14, B=\\dfrac{6}{11}"

Then


"y_p=-14+\\dfrac{6}{11}e^{2x}"

The general solution of the given non-homogeneous differential equation is


"y=y_h+y_p"




"y=C_1e^{(-2-\\sqrt{5})x}+C_2e^{(-2+\\sqrt{5})x}-14+\\dfrac{6}{11}e^{2x}"


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