Can the Separable First Order ODE be solved with a different answer?

In summary: So it checks out.In summary, the original equation 2r(s^2+1)dr + (r^4 + 1)ds = 0 can be rewritten as r² + s = c(1 - r² s), which can be derived using the addition theorems for sine and cosine. The solution involves using the tangent of both sides and substituting c to get (1 + s²) 2r dr + (1 + r⁴) ds = 0, which matches the original equation and checks out.
  • #1
shelovesmath
60
0
2r(s^2+1)dr + (r^4 + 1)ds = 0
2.book answer different than mine...book's answer: r^2 + s = c(1 -r^2 s)
3. 2r(s^2+1)dr =- (r^4 + 1)ds
-2r/r^4+1 dr = 1/s^2+1 ds
int -2r/r^4+1 dr = int 1/s^2+1 ds
with u substitution on left we have u = -2r, etc.
tan^-1 r^2 = - tan^ -1 s + c
tan^-1 r^2 + tan^-1 s = c
 
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  • #2
Don't forget brackets in writing, it may confuse people.
We have ∫ -2r/(r⁴+1) dr = ∫ 1/(s²+1) ds. The left integral gives -arctan(r²), the right arctan(s), so c - arctan(r²) = arctan(s). (You forgot the minus sign from the variable change). Taking the tangens on both sides and using the addition theorem we have
s = tan [ c - arctan(r²) ] = ( tan c - r² )/( 1 + r² tan c ), so that with c' = tan c we get
( 1 + r² c' ) s = c' - r² which is exactly what your book gave you.
 
  • #3
grey_earl said:
Taking the tangens on both sides and using the addition theorem we have
s = tan [ c - arctan(r²) ] = ( tan c - r² )/( 1 + r² tan c ), so that with c' = tan c we get
( 1 + r² c' ) s = c' - r² which is exactly what your book gave you.

I, uh, don't remember the tangents theorem. Calculus topic? Precalc? I'll go google I guess. Thanks for the feedback.
 
  • #4
The derivative of the answer should then match the initial problem; however, this did not work out for me. Does it check out for you?
 
  • #5
It doesn't look as you have completed the calculation. Take tan of both sides and use the trig expression for tan(A+B)
 
  • #6
We have tan(a+b) = [ tan a + tan b ]/[ 1 - tan a tan b ], which can be derived using the addition theorems for sine and cosine.

And, if you derive r² + s = c(1 - r² s), you should get 2 r dr + ds = - c (2 r dr s) - c (r² ds), so ( 1 + c s ) 2 r dr + ( 1 + c r² ) ds = 0, which after substituting c from the result and multiplying by (1- r² s) simplifies to ( 1 + s² ) 2 r dr + ( 1 + r⁴ ) ds = 0, which is the original equation.
 
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What is a separable first order ODE?

A separable first order ODE is a type of ordinary differential equation (ODE) that can be written in the form dy/dx = f(x)g(y), where f(x) and g(y) are functions of x and y, respectively. This means that the differential equation can be separated into two parts, one involving only x and the other only involving y.

How do you solve a separable first order ODE?

To solve a separable first order ODE, you first need to separate the variables by moving all terms involving y to one side and all terms involving x to the other side. Then, you can integrate both sides with respect to x and y separately to get the general solution. Finally, you can use initial conditions to find the particular solution.

What are the applications of separable first order ODEs?

Separable first order ODEs have many applications in physics, engineering, and other scientific fields. They are commonly used to model growth and decay processes, population dynamics, and chemical reactions. They are also used in circuit analysis, mechanics, and fluid dynamics.

What are the limitations of separable first order ODEs?

One of the main limitations of separable first order ODEs is that they can only be used to solve certain types of ODEs. They cannot be used to solve higher order ODEs or systems of ODEs. Additionally, the method of separation of variables may not always work for more complex ODEs.

Can separable first order ODEs have multiple solutions?

Yes, separable first order ODEs can have multiple solutions. However, these solutions may not always be valid or physically meaningful. It is important to check the solutions obtained by solving the ODE to ensure they satisfy any given initial conditions and make sense in the context of the problem being studied.

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