Find the solution of the differential equation # x(1+y^2)dx-y(1+x^2)dy =0#?

1 Answer
Oct 6, 2017

# y^2 = ( x^2 -1)/2 #

Explanation:

We have:

# x(1+y^2)dx-y(1+x^2)dy =0#

If we rearrange this ODE from differential form into standard form we have:

# y(1+x^2)dy/dx =x(1+y^2) #

# :. y/(1+y^2)dy/dx =x/(1+x^2) #

This is now a separable ODE, do we can "separate the variables" to give:

# int \ y/(1+y^2)dy = int \ x/(1+x^2) \ dx#

We can manipulate the numerators on both sides to get:

# int \ (2y)/(1+y^2)dy = int \ (2x)/(1+x^2) \ dx#

Both integrals are identical and of standard functions, so we can now integrate to get the General Solution:

# ln | 1+y^2 | = ln | 1+x^2 | + C #

We are given that #y=0# when #x=1#, so:

# ln | 1 | = ln | 1+1 | + C +> C = -ln2#

So we have the Particular Solutions

# ln | 1+y^2 | = ln | 1+x^2 | -ln2 #

# :. ln ( 1+y^2 ) = ln (( 1+x^2 )/2 ) #

# :. 1+y^2 = ( 1+x^2 )/2 #

# :. y^2 = ( 1+x^2 )/2 -1 #

# :. y^2 = ( x^2 -1)/2 #