1. ## Integration need help

Solve $\displaystyle (x^2+1)y' + y^2 + 1 =0, y(3)=2$. Find y in terms of x only. Simplify so that your answer does not contain any trigonometric or inverse trig functions.

$\displaystyle (x^2+1)y' + y^2 + 1 =0, y(3)=2$
$\displaystyle dy/dx + y^2 = -\frac{1}{x^2+1}$

I'm stuck here, how do I get it so I can have y^2 multiplied by dy and have the other side multiplied by dx so I can integrate?

2. You should end up here...

$\displaystyle dy/dx= -\frac{y^2+1}{x^2+1}$

since $\displaystyle (x^2+1)$ is only factor for y' and not $\displaystyle y^2$.

Go on from there...

3. ok so when you integrate, on the left side u get y, but how do u deal with the right side since there's both x and y?

4. You have a separable DE.

$\displaystyle \left( {x^2 + 1} \right)y' + \left( {y^2 + 1} \right) = 0 \implies \left( {x^2 + 1} \right)\,\frac{{dy}} {{dx}} = - \left( {y^2 + 1} \right)$

so we have $\displaystyle - \frac{1} {{y^2 + 1}}\,dy = \frac{1} {{x^2 + 1}}\,dx$

Now it remains to integrate both sides.

5. Originally Posted by circuscircus
ok so when you integrate, on the left side u get y, but how do u deal with the right side since there's both x and y?
I only intended to make you aware of a mistake during your first steps in rearranging the factors...