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Find the general solution of the differential equation x^2y′′+xy′−y=2x for x>0 by first showing that the function yp=−12x+x ln(x)is a particular solution for the differential equation.

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nikita83
[tex]x^2y''+xy'-y &=2x\\ x^2y''+2xy'-xy'-y &=2x\\ (x^2y')'-(xy)' &=2x\\ (x^2y'-xy)' &=2x\\ x^2y'-xy &=x^2+C_1\\ xy'-y &=x+\frac{C_1}{x}\\ x^2\left(\frac{y}{x}\right)' &=x+\frac{C_1}{x}\\ \left(\frac{y}{x}\right)' &=\frac{1}{x}+\frac{C_1}{x^3}\\ \frac{y}{x} &=\ln{x}-\frac{C_1}{2x^2}+C_2\\ y &=x\ln{x}-\frac{C_1}{2x}+C_2x[/tex]

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