Is My Solution to the Reduction of Order Problem Correct?

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SUMMARY

The discussion focuses on solving the reduction of order problem for the differential equation \(x^2y'' + 3ty' + y = 0\) given the first solution \(y_1 = x^{-1}\). The user applied Abel's method and derived the second solution as \(y_2 = \frac{\ln(x)}{x}\). The correctness of this solution is questioned, prompting requests for detailed steps to verify the approach.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with the reduction of order technique in solving differential equations.
  • Knowledge of Abel's method for finding solutions to linear differential equations.
  • Basic calculus, including logarithmic functions and their properties.
NEXT STEPS
  • Review the application of Abel's method in solving linear differential equations.
  • Practice solving reduction of order problems with different first solutions.
  • Explore the verification of solutions using substitution back into the original differential equation.
  • Study the properties of logarithmic functions in the context of differential equations.
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Students studying differential equations, mathematics educators, and anyone looking to enhance their problem-solving skills in advanced calculus topics.

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I think I have this right, its just its an online assignment and I don't want to lose any points on it.(1/4 my grade)

Question: Find y2 if y1= x^(-1) and the equation: x^2y"+3ty'+y=0.

I solved(using Abel's) for it:
ln(x)/x

but I am not entirely sure.

Wondering if I am spot on or am not doing it right/ thanks
 
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bigt9 said:
I think I have this right, its just its an online assignment and I don't want to lose any points on it.(1/4 my grade)

Question: Find y2 if y1= x^(-1) and the equation: x^2y"+3ty'+y=0.

I solved(using Abel's) for it:
ln(x)/x

but I am not entirely sure.

Wondering if I am spot on or am not doing it right/ thanks

Welcome to the PF.

Can you please show the steps you used to arrive at that solution?
 

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