Solving a second order ODE using reduction of order

Click For Summary
SUMMARY

The discussion focuses on solving a second-order ordinary differential equation (ODE) using the method of reduction of order. The user encountered issues with their solution, specifically with the verification step where the left-hand side (LHS) and right-hand side (RHS) did not match. A key mistake identified was the incorrect sign in the derivative, where the user incorrectly wrote ##y'=-\frac{1}{2} t^{-\frac{1}{2}}## instead of the correct ##y'=\frac{1}{2} t^{-\frac{1}{2}}## for the solution ##y=t^{\frac{1}{2}}##. This correction clarified the user's misunderstanding and confirmed the validity of the general solution provided by Wolfram Alpha.

PREREQUISITES
  • Understanding of second-order ordinary differential equations (ODEs)
  • Familiarity with the method of reduction of order
  • Basic knowledge of derivatives and their notation
  • Experience with verification of solutions in differential equations
NEXT STEPS
  • Study the method of reduction of order in detail
  • Practice solving various second-order ODEs
  • Learn how to verify solutions for differential equations
  • Explore computational tools like Wolfram Alpha for solving ODEs
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, as well as anyone seeking to improve their problem-solving skills in ODEs.

Bonnie
Messages
18
Reaction score
1

Homework Statement


Hi there, I have an assignment which involves using reduction of order to solve for a second solution to an ode (the one attached). However this is a method I am new to, and though I have tried several times, I'm somehow getting something wrong because the LHS and RHS are not matching up, that is, when I substitue in the solution I have found, the RHS does not equal zero as it should.

Homework Equations

The Attempt at a Solution


I have attached my working (Sides 1 and 2), if anyone could point out what I'm doing wrong it would be greatly appreciated, this is driving me nuts!
 

Attachments

  • Question.JPG
    Question.JPG
    8.4 KB · Views: 443
  • s1.jpg
    s1.jpg
    54.4 KB · Views: 481
  • s2.jpg
    s2.jpg
    62.3 KB · Views: 441
Physics news on Phys.org
According to wolfram your general solution is correct.

You do a blunt mistake in the check, you put ##y'=-\frac 1 2 t^{-\frac 1 2}##, the minus in front is not needed. It is clear that ##y'=\frac{1}{2} t^{-\frac 1 2}## for ##y=t^{\frac 1 2}##
 
  • Like
Likes   Reactions: SammyS
Delta² said:
According to wolfram your general solution is correct.

You do a blunt mistake in the check, you put ##y'=-\frac 1 2 t^{-\frac 1 2}##, the minus in front is not needed. It is clear that ##y'=\frac{1}{2} t^{-\frac 1 2}## for ##y=t^{\frac 1 2}##
Oh, that was dumb. Thank you!
 
  • Like
Likes   Reactions: Delta2

Similar threads

Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 33 ·
2
Replies
33
Views
4K
Replies
7
Views
2K
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
Replies
7
Views
2K