[SOLVED] Differential Equation

• Feb 3rd 2009, 06:01 PM
ronaldo_07
[SOLVED] Differential Equation
Determine the general solution of the exact differential equation
1− $\frac{x}{x^2+y^2}$-( $\frac{y}{x^2+y^2}$) $\frac{dy}{dx}$=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?
• Feb 3rd 2009, 06:03 PM
Mush
Quote:

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

$1 - \frac{x}{x^2+y^2} - \bigg(\frac{y}{x^2+y^2}\bigg)\frac{dy}{dx} = 0$?
• Feb 3rd 2009, 06:19 PM
ronaldo_07
yes thats the question
• Feb 8th 2009, 05:27 PM
ronaldo_07
Any help on solving this anyone?
• Feb 8th 2009, 05:36 PM
Mush
Quote:

Originally Posted by ronaldo_07
Any help on solving this anyone?

Multiply through by $dx$ which gives:

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

You now have it in the form:

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

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

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

You should get a solution of:

$x - y + x\arctan\bigg(\frac{y^2}{x^2}\bigg) = c$
• Feb 8th 2009, 05:37 PM
Chris L T521
Quote:

Originally Posted by Mush
Do you mean

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

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

Can you try to take it from here?