Differential equations - solving initial value problem

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SUMMARY

The discussion focuses on solving the initial value problem defined by the differential equation y' = -4t/y with the initial condition y(0) = y0. The solution derived is y = +/- sqrt((y0)^2 - 4t^2), indicating that the solution exists as long as the expression under the square root remains non-negative. This leads to the conclusion that the interval of existence for the solution depends on the initial value y0, specifically that the solution is valid for t in the range of -y0/2 to y0/2, ensuring the square root remains real-valued.

PREREQUISITES
  • Understanding of differential equations, specifically first-order separable equations.
  • Knowledge of initial value problems and their significance in differential equations.
  • Familiarity with the concept of real-valued functions and conditions for their existence.
  • Basic algebraic manipulation, particularly with square roots and inequalities.
NEXT STEPS
  • Study the properties of first-order differential equations and their solutions.
  • Learn about the existence and uniqueness theorem for initial value problems.
  • Explore the implications of initial conditions on the behavior of solutions to differential equations.
  • Investigate other methods for solving differential equations, such as numerical approaches or qualitative analysis.
USEFUL FOR

Students and educators in mathematics, particularly those focusing on differential equations, as well as anyone looking to deepen their understanding of initial value problems and their solutions.

DWill
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Homework Statement


Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y0.

y' = -4t/y, y(0) = y0


Homework Equations





The Attempt at a Solution


I solved the DE and got to:

y = +/- sqrt(C - 4t^2)

Plugging in y_0 for y when t = 0, I get:

y = +/- sqrt( (y_0)^2 - 4t^2 )

I'm pretty sure this is the right, but how do I answer the last part of the question? Figuring out the how the interval depends on initial value? thanks
 
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DWill said:

Homework Statement


Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y0.

y' = -4t/y, y(0) = y0


Homework Equations





The Attempt at a Solution


I solved the DE and got to:

y = +/- sqrt(C - 4t^2)

Plugging in y_0 for y when t = 0, I get:

y = +/- sqrt( (y_0)^2 - 4t^2 )

I'm pretty sure this is the right, but how do I answer the last part of the question? Figuring out the how the interval depends on initial value? thanks
Usually, unless stated otherwise, one would assume that the solution must be real-valued.
 

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