First order differential equation question

Click For Summary
SUMMARY

The discussion centers on solving the first-order differential equation dy/dx=(x(x^2+1))/4y^3 with the initial condition y(0)=-1/√2. The user initially integrates both sides correctly but arrives at a different solution than the textbook, which states y=-(√(x^2+2)/2). The discrepancy arises from a potential error in the integration process and the application of the boundary condition. The advice given includes revisiting the integration steps and being cautious with algebraic simplifications and root extractions.

PREREQUISITES
  • Understanding of first-order differential equations
  • Proficiency in integration techniques
  • Familiarity with boundary value problems
  • Knowledge of algebraic manipulation and simplification
NEXT STEPS
  • Review integration techniques for first-order differential equations
  • Study the application of boundary conditions in differential equations
  • Learn about algebraic simplification methods in calculus
  • Explore the use of Mathematica for solving differential equations
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, as well as anyone using computational tools like Mathematica for solving complex mathematical problems.

idks16
Messages
1
Reaction score
0
The problem is : dy/dx=(x(x^2+1))/4y^3 when y(0)=-1/√2
This is my work so far:
∫4y^3dy=∫x(x^2+1)dx
(y^4)/2=((x^2+1)^2)/2+c
The answer from the textbook is y=-(√(x^2+2)/2)
As you can see, my work will never equal the textbook answer when you put it in the y= stuff form. What did I do wrong?
 
Physics news on Phys.org
I got a slightly different answer than what you posted from the text

y(x) = -\sqrt{\frac{1}{2}(x^2+1)}

and mathematica agrees with me, so perhaps a typo?

Anyway, it looks like your on the right track, although go back through the integration, I think you may be off by a factor.
Then apply the boundary condition to find the integration constant.
And simplify the algebra down to the answer.
Also be conscience of taking roots,
y(x) = ±(stuff)^{1/4}
good luck
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
18
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 52 ·
2
Replies
52
Views
8K