Easy ODE question, but I'm missing something

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In summary, the conversation discusses solving the equation (ds/dy)=(y+2s)/(y-2s) using the substitution V=s/y and transforming it into a separable equation. The speaker has trouble integrating the right side but is reminded that the denominator can be rewritten as a perfect square. The final solution involves setting u= V+ (1/4).
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Homework Statement



(ds/dy)=(y+2s)/(y-2s)

Homework Equations





The Attempt at a Solution



I let V= s/y and this gave me *(ds/dy)=(1+2V)/(1-2V)

Then because V= s/y I said s=Vy and (ds/dy)=V + y(dV/dy)

Right so then my equation looked like V + y(dV/dy)=(1+2V)/(1-2V)

and then obviously I could do this : y(dV/dy)=(1+2V)/(1-2V) - V

then because I wanted to make V part of the quotient so It becomes
(1+V+2V^2)/(1-2V) = y(dV/dy)

now this is a separable equation, so I can transform it to
(dy/y)= (1-2V)(dV)/(1+V+2V^2) so, integrating both sides gives me LN|y|= the integral of the right side, but, how do you do this? this is where I get lost. I think the answer might be that (1+2V)/(1-2V) can be re written in a simplified way buit I can't remember it. Plz help.
 
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  • #2
Your denominator, 2V2+ V+ 1 does not have real zeroes and so cannot be factored in terms of real numbers. What you can do is complete the square: 2V2+ V+ 1= 2(V2+ (1/2)V)+ 1= 2(V2+ (1/2)V+ (1/16)- (1/16)+ 1= 2(V+ (1/4))2- 1/32+ 1= 2(V+ (1/4))2+ 31/32 so (1- 2V)/(1+ V+ 2V2)= (1/2)(1-2V)/((V+(1/4))2+ 31/32). Now let u= V+ (1/4).
 
  • #3
thanks alot.
 

FAQ: Easy ODE question, but I'm missing something

What is an ODE?

An ODE, or ordinary differential equation, is a type of mathematical equation that describes the relationship between a function and its derivatives. It is used to model real-world systems that involve continuous change over time.

How do you solve an ODE?

ODEs can be solved analytically or numerically. Analytical solutions involve finding an exact mathematical expression for the solution, while numerical solutions involve using approximation methods to find an approximate solution.

What are the applications of ODEs?

ODEs have numerous applications in various fields such as physics, engineering, economics, and biology. They are used to model phenomena such as population growth, chemical reactions, and electrical circuits.

What are initial conditions in ODEs?

Initial conditions are the values of the dependent variable and its derivatives at a specific initial time. They are needed to find a unique solution to an ODE.

What are some common techniques for solving ODEs?

Some common techniques for solving ODEs include separation of variables, integrating factors, and series solutions. Numerical methods such as Euler's method and Runge-Kutta methods are also commonly used.

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