Why Does My Differential Equation Solution Differ by a Sign?

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SUMMARY

The discussion centers on the discrepancy between a user's solution to a differential equation and the textbook answer, which differs by a negative sign. The user is reassured that their solution is correct as it represents the general solution of the equation. The uniqueness of the solution is established through the application of initial conditions, such as y(0)=y_0, which allows for the adjustment of integration constants to align with the textbook's answer. The conversation highlights the importance of understanding integration constants in differential equations.

PREREQUISITES
  • Understanding of differential equations and their general solutions.
  • Familiarity with initial value problems and constraints.
  • Knowledge of integration constants and their role in solutions.
  • Ability to differentiate functions to verify solutions.
NEXT STEPS
  • Study the concept of uniqueness in differential equations.
  • Learn about initial value problems and their significance in determining solutions.
  • Explore the role of integration constants in differential equations.
  • Practice verifying solutions by differentiation and substitution.
USEFUL FOR

Students and educators in mathematics, particularly those studying differential equations, as well as anyone seeking to clarify the concept of integration constants and initial conditions in mathematical solutions.

rxfudd
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I cannot figure out where I am going wrong. My answer and the textbook answer are different by a negative sign. Can someone review my work and tell me what I am doing wrong?
 

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Why do you think, your solution is wrong? It's perfectly right! It's the general solution of the equation (I've not checked it, but this you can do by taking the derivative and check whether it fulfills your equation), and whether you call the integration constant -C_3 or C doesn't matter.

The solution is made unique by giving, e.g., an initial value y(0)=y_0 as a constraint. If you use this in your and the textbook's answer you'll get the same result by choosing the right C_3 and C to match the initial-value constraint, respectively.
 
Ah yes, the constant...the (x2+4)-4 term can be positive or negative because of the constant.

Thank you for reviewing my work and pointing that out. Amateur mistake :)
 

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