Solving a Nasty First Order ODE

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SUMMARY

The discussion focuses on solving a first-order ordinary differential equation (ODE) from Blanchard's "Differential Equations," specifically question 32. The equation presented is \(\frac{d}{dt}y(t) - \frac{y(t)t^3}{1+t^4} = 2\). The initial attempt using an integrating factor was deemed unworkable, leading to suggestions for a simpler approach involving multiplying through by \((1+t^4)\) to facilitate the application of the Quotient Rule. The key insight is recognizing the derivative of \((1+t^4)\) to simplify the left-hand side.

PREREQUISITES
  • Understanding of first-order ordinary differential equations
  • Familiarity with integrating factors in ODEs
  • Knowledge of the Quotient Rule in calculus
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the method of integrating factors for first-order ODEs
  • Learn about the Quotient Rule and its applications in differential equations
  • Practice solving first-order ODEs using various techniques
  • Explore examples from Blanchard's "Differential Equations" for deeper understanding
USEFUL FOR

Students studying differential equations, educators teaching calculus concepts, and anyone seeking to improve their problem-solving skills in first-order ODEs.

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Homework Statement



This problem is from Blanchard "Differential Equations" Chapter one review, question 32.

{\frac {d}{dt}}y \left( t \right) -{\frac {y \left( t \right) {t}^{3}}<br /> {1+{t}^{4}}}=2<br />


The Attempt at a Solution



Using an integrating factor yields:

<br /> {\frac {d}{dt}} \sqrt [4]{1+{t}^{4}}y \left( t \right) =2\,\sqrt [4]{1+{t}^{4}}<br />

This is unworkable...

Any hints would be greatly appreciated. I suspect that I am overlooking a simple guess.
 
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Yeah, you're making it much harder than it is. Just multiply out by (1+t^4)dt, and you get

(1+t^4)dy - yt^3dt = 2(1+t^4)dt

Hint: try to make the left side look like the Quotient Rule. What is the derivative of (1+t^4)?
 

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