How Do You Isolate Variables in a Complex Differential Equation?

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SUMMARY

The discussion focuses on solving the differential equation m \ddot{y} = -mg - \beta \dot{y}. The integration process leads to the first-order equation \frac{dy}{dt} = -gt - \frac{\beta y}{m} + C. Participants suggest isolating variables by letting v = dy/dt and transforming the equation into m \frac{dv}{dt} = -mg - \beta v. The solution for v is derived as v = C e^{-\frac{\beta}{m}t} - \frac{mg}{\beta}, which ultimately allows for the integration to find y(t), involving two constants.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with integration techniques and logarithmic functions.
  • Knowledge of variable substitution methods in calculus.
  • Basic grasp of constants and their roles in differential equations.
NEXT STEPS
  • Study the method of integrating factors for first-order differential equations.
  • Learn about the Laplace transform and its applications in solving differential equations.
  • Explore the concept of initial value problems and their solutions in differential equations.
  • Investigate the use of numerical methods for solving complex differential equations.
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Students, mathematicians, and engineers who are working with differential equations and require techniques for isolating variables and solving complex equations.

stunner5000pt
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can you help me olve this (simple?) differential equation
[tex]m \ddot{y} = -mg - \beta \dot{y}[/tex]
integrate once and i get
[tex]\dot{y} = -gt - \frac{\beta y}{m} + C[/tex]
also can be written as
[tex]\frac{dy}{dt} = -gt - \frac{\beta y}{m} + C[/tex]
both are equivalent

basically trying to get the y on one side adn the t on the other side. HJva tried many ways but can't isolate the two. Any suggestions?
 
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Have you tried the approach where you assume a solution for y(t), differentiate it twice and work out the equation constants...? What would be a typical function that could work with this approach...?
 
First of all let v = dy/dt, then you have
[itex]m \frac{dv}{dt} = -mg - \beta v[/itex]
From which

[itex]\frac{dv}{g + \frac{\beta}{m} v} = -dt[/itex]

Integrating it now gives

[itex]\frac{m}{\beta} \ln(g + \frac{\beta}{m} v) = -t + C_1[/itex]

Solving for v gives:

[itex]v = C e^{-\frac{\beta}{m}t} - \frac{mg}{\beta}[/itex]

Where C is some new constant which is I think [itex]C = \frac{m}{\beta}e^{\frac{\beta}{m} C_1}[/itex]. But it doesn't matter though.

Now remember that v = dy/dt. Thus integrating the last equation will give you y(t). (there will be two constants then)
 
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