# Thread: [SOLVED] Another Separable ODE

1. ## [SOLVED] Another Separable ODE

Ok another one

dy/dx = 2y/1- x^4

sol: y = C|1+x/1-x|^1/2 e^arctanx

Very confusing this is how i start out:

=> |2y dy = | 1/1-x^4 dx

=> y^2 = 1/4 lnu

=> y^2 = 1/4 ln(1-x^4) +c

=> y = (1/4 ln(1-x^4)+c )^1/2

then im lost, and ive probably stuffed it up right from near the start. im not sure if i should be using integration by substitution when integrating the x side.
Help appreciated

2. Originally Posted by sterps
Ok another one

dy/dx = 2y/1- x^4

sol: y = C|1+x/1-x|^1/2 e^arctanx

Very confusing this is how i start out:

=> |2y dy = | 1/1-x^4 dx

=> y^2 = 1/4 lnu

=> y^2 = 1/4 ln(1-x^4) +c

=> y = (1/4 ln(1-x^4)+c )^1/2

then im lost, and ive probably stuffed it up right from near the start. im not sure if i should be using integration by substitution when integrating the x side.
Help appreciated
You've made a number of errors. Note: $\int \frac{dy}{2y} = \int \frac{dx}{1 - x^4}$.

I suggest partial fractions for the integral on the right hand side. Use the decomposition $\frac{1}{1 - x^4} = \frac{A}{1 - x} + \frac{B}{1 + x} + \frac{Cx + D}{1 + x^2}$.

3. Originally Posted by sterps

=> |2y dy = | 1/1-x^4 dx

=> y^2 = 1/4 lnu

=> y^2 = 1/4 ln(1-x^4) +c

=> y = (1/4 ln(1-x^4)+c )^1/2
1/4 ln(1-x^4) +c is not anti-derivative of 1/1-x^4

Do you know partial fraction?

EDIT : wew, a little bit late. Mr_fantastic has given the clue ^^

4. Originally Posted by sterps
Ok another one

dy/dx = 2y/1- x^4

If you are going to post you question in plain ASCII use sufficient brackets to render you meaning unambiguous:

dy/dx = 2y/(1- x^4)

would do here.

CB

5. $\frac{1}{1-x^{4}}=\frac{1}{\left( 1-x^{2} \right)\left( 1+x^{2} \right)}=\frac{1}{2}\cdot \frac{\left( 1+x^{2} \right)+\left( 1-x^{2} \right)}{\left( 1-x^{2} \right)\left( 1+x^{2} \right)}.$