Answer to Question #164164 in Differential Equations for mehro

Question #164164

d^3y/dx^3-3 d^2y/dx^3+ 4 dy/dx -2y=e^x+cosx


1
Expert's answer
2021-02-16T13:53:22-0500

Solution.

"\\frac{d^3y}{dx^3}-3\\frac{d^2y}{dx^2}+4\\frac{dy}{dx}-2y=e^x+\\cos\u2061{x},"

or


"y'''-3y''+4y'-2y=e^x+\\cos{x}."


This is a linear differential equation of the third-order with constant coefficients. The general solution of this equation is found as the sum of the general solution "\\tilde{y}" of the corresponding homogeneous equation and some particular integral solution "y_*" of the inhomogeneous equation: "y=\\tilde{y}+y_*."

Сompose and solve the characteristic equation:


"\\lambda^3-3\\lambda^2+4\\lambda-2=0,\n\\newline(\\lambda-1)(\\lambda^2-2\\lambda+2)=0,\n\\newline\n\\lambda_1=1, \\text{or } \\lambda^2-2\\lambda+2=0,\n\\newline \\text{ } \n\\lambda_2=1+i, \\text{ } \\lambda_3=1-i."


Therefore, "\\tilde{y}=C_1e^{\\lambda_1x}+C_2e^{ax}\\cos{bx}+C_3e^{ax}\\sin{bx},"

where "a" is a real part, and "b" is an imaginary part of complex numbers "\\lambda_2 \\text{ and } \\lambda_3." So,


"\\newline \\tilde{y}=C_1e^x+C_2e^x\\cos{x}+C_3e^x\\sin{x}."

The particular integral solution "y_*" is writte in the form:


"y_\u2217=y_1+y_2\u200b."

Since the right-hand side contains a "\\cos{x}" , we look for a particular integral solution "y_1" ​in the form:


"y_1=A\\cos{x}+B\\sin{x}."

Using the method of undetermined coefficients we find that "A=\\frac{1}{10}" and "B=\\frac{3}{10}." So, "y_1=\\frac{1}{10}\\cos{x}+\\frac{3}{10}\\sin{x}."

Since the right-hand side also contains a "e^{x}" and 1 is characteristic number, we find a particular integral solution "y_2" ​in the form:


"y_2=Cxe^{x}."

Using the method of undetermined coefficients we find that "C=1."

Therefore, "y_2=xe^{x}."

As follows, the particular integral solution of "\\frac{d^3y}{dx^3}-3\\frac{d^2y}{dx^2}+4\\frac{dy}{dx}-2y=e^x+\\cos\u2061{x}," is


"y_*=\\frac{1}{10}\\cos{x}+\\frac{3}{10}\\sin{x}+xe^{x},"

and the general solution of the same equation is


"y=C_1e^x+C_2e^x\\cos{x}+C_3e^x\\sin{x}+xe^x+\\frac{1}{10}\\cos{x}+\\frac{3}{10}\\sin{x}."

Answer.

"y=C_1e^x+C_2e^x\\cos{x}+C_3e^x\\sin{x}+xe^x+\\frac{1}{10}\\cos{x}+\\frac{3}{10}\\sin{x}."



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