1. ## [SOLVED] Differential Equation

Determine the general solution of the exact differential equation
1−$\displaystyle \frac{x}{x^2+y^2}$-($\displaystyle \frac{y}{x^2+y^2}$)$\displaystyle \frac{dy}{dx}$=0

Fix the constant of integration according to the initial condition y(0) = e and write down the explicit form of the solution.

How do I solve this question?

2. Originally Posted by ronaldo_07
Determine the general solution of the exact differential equation
1−$\displaystyle \frac{x}{x^2+y^2}$-($\displaystyle \frac{y}{x^2+y^2}$)$\displaystyle \frac{dy}{dx}$=0

Fix the constant of integration according to the initial condition y(0) = e and write down the explicit form of the solution.

How do I solve this question?
Do you mean

$\displaystyle 1 - \frac{x}{x^2+y^2} - \bigg(\frac{y}{x^2+y^2}\bigg)\frac{dy}{dx} = 0$?

3. yes thats the question

4. Any help on solving this anyone?

5. Originally Posted by ronaldo_07
Any help on solving this anyone?
Multiply through by $\displaystyle dx$ which gives:

$\displaystyle (1-\frac{x}{x^2+y^2})dx - \bigg(\frac{y^2}{x^2+y^2}\bigg) dy = 0$

You now have it in the form:

$\displaystyle f(x,y)dx + g(x,y)dy = 0$

The general solution to an equation of this form is given by:

$\displaystyle \int f(x,y)\partial x + \int \bigg(g(x,y) - \frac{\partial}{\partial y} \int f(x,y)dx \bigg) \partial y = c$

You should get a solution of:

$\displaystyle x - y + x\arctan\bigg(\frac{y^2}{x^2}\bigg) = c$

6. Originally Posted by Mush
Do you mean

$\displaystyle 1 - \frac{x}{x^2+y^2} - \bigg(\frac{y}{x^2+y^2}\bigg)\frac{dy}{dx} = 0$?
Multiply both sides by $\displaystyle x^2+y^2$ to get the equation $\displaystyle x^2+y^2-x-y\frac{\,dy}{\,dx}=0\implies \frac{\,dy}{\,dx}-y=\left(x^2-x\right)y^{-1}$. This is a Bernoulli equation.

Can you try to take it from here?