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

[shelovesmath]

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

 

[grey_earl]

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.

 

[shelovesmath]

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.

 

[jonspalding]

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?

 

[hunt_mat]

It doesn't look as you have completed the calculation. Take tan of both sides and use the trig expression for tan(A+B)

 

[grey_earl]

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.