Solving a Differential Equation: 2y(1+x^2)y' + x(1+y^2) = 0

[ProPatto16]

Homework Statement

Solve the differential equation 2y(1 + x2 )y'+ x(1 + y2 ) = 0 where y = 2 when x = 0.
given answers are
a. (1 + x2 )(1 + y2 )2= 0
b. (1 + x2 )2 (1 + y2 ) = 25
c. (1 + x2 )2 (1 + y2 )2= 0
d. (1 + x2 )(1 + y2 ) = 0
e. (1 + x2 )(1 + y2 )2= 25

The Attempt at a Solution

2y(1+x²)dy/dx+x(1+y²)=0
2y(1+x²)dy/dx=-x(1+y²)
2y/(1+y²)dy = -x/(1+x²)dx

if i integrate these i end up with ln in the equations. so i don't know how to do it so it relates to the given answers?

 

[phyzguy]

Exponentiate both sides to get rid of the log, and you should get one of the answers.

 

[ProPatto16]

if i keep going from where i left off.. after integrating you get:

ln(1+y²) = - ln(1+x²)/2

then taking off log of both sides leaves

(1+y²) = -(1+x²)/2

take all onto one side gives:

2(1+y²)+(1+x²)

where did i go wrong?

 

[HallsofIvy]

If you are given answers to select from, you don't need to integrate at all. Just put each of the possible answer into the equation and see if it satisfies it!

"Taking the log off" both sides of $ln(1+y^2)= ln(1+ x^2)/2$ does NOT give "$1+ y^2= (1+ x^2)/2$: the "2" in the denominator is not inside the logarithm. Instead, write it as $ln(1+ y^2)= ln((1+ x^2)^{1/2})$ so that $1+ y^2= (1+ x^2)^{1/2}$ or $(1+ y^2)^2= 1+ x^2$.

 

[ProPatto16]

haha i have to show how i got every answer, mathematically. otheriwse yeah id trial and error the questions i couldn't do.

the denominator there is where i thought i went wrong.

i think i know the answer but its that one sign i can't get rid of.

from the start:

2y(1+x2)dy/dx+x(1+y2)=0
2y(1+x2)dy/dx=-x(1+y2)
2y/(1+y2)dy = -x/(1+x2)dx
then integrate
ln(1+y2) = - ln(1+x2)/2
which becomes
2ln(1+y2) = - ln(1+x2)
take 2 and into logarithm
ln((1+y2)²) = - ln(1+x2)

waittt... ifi take -1 into logarithm then i get an inverse funtion, which will get rid of the plus/minus sign...

then sub in y=2 and x=0


Thanks!