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Two linearly independent solutions $y_{1}$ and $y_{2}$ are given to the associated homogeneous equation of the variable-coefficient nonhomogeneous equation. Use the method of variation of parameters to find a particular solution to the nonhomogeneous equation. Assume $x>0$ in each exercise.$$x^{2} y^{\prime \prime}+x y^{\prime}-y=x, \quad y_{1}=x^{-1}, y_{2}=x$$

Calculus 3

Chapter 17

Second-Order Differential Equations

Section 2

Nonhomogeneous Linear Equations

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Okay, so our particular solution is going to have the form of u of x, times x, plus b of x times x, to negative 2 were not? U, prime of x, plus v prime of x, negative 2 is equal to 0 and u prime minus 2 v. Prime x to negative 3 is equal to 3 x, 2 negative 1 n. So we have that: u, prime xte, minus 2 v prime, is equal to 3 x, squared okay. If we multiply this by 2 x, cube and 3 by x, and then some of them we'll forget: 2- u prime x, to the 4 plus 2 v prime x plus: u prime x, 4 minus 2 x v prime isutthree x, cube, which simplifies to u prime X, 4 plus 2 x 4 is equal to 3 x cubed. So we get the, u prime, is equal to 1 over x and then, if we take the integral both sides, we get that. U is equal to l n of x now if we multiply equation 2 by negative x squared and that's your equation, this equation here multiply it by negative x squared and i, on this 1. We get that: u, prime negative x, cube minus v, prime plus. U prime b, cube or x cubed minus 2 t prime, is equal to 3 x squared. So we get that the prime is equal to negative x squared in a booth sides. We get. That b is equal to negative x cubed over our particular solution becomes x. L n of xx over 3.

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