# Thread: Solve first order ODE

1. ## Solve first order ODE

The ODE is
$xdx+ydy=\frac {a^2 (xdy-ydx)}{x^2+y^2}$

We can take $x^2+y^2$ on the left hand side to get
$(x^2+y^2)d (x^2+y^2)=2a^2 (xdy-ydx)$

I am not able to move ahead from here.

2. ## Re: Solve first order ODE

The first thing to do would be to gather all the similar differentials together.

$\big((x^2+y^2)x+2a^2y)\mathrm dx + \big((x^2+y^2)y-2a^2x\big)\mathrm dy)=0$

I would then test to see if it is an exact equation. If it isn't, seek an integrating factor.

3. ## Re: Solve first order ODE

I have tried doing this. The problem here is that the DE is not exact and you can only use the integration factor 1/bx+ay where b is the eqn before dx and a is eqn before dy.The integration factor results in an equation with more than 1 term which makes figuring out the differential very unobvious and time consuming.

4. ## Re: Solve first order ODE

On the contrary, the equation you get by multiplying both sides by $x^2+ y^2$ is exact.

$(x^2+ y^2)(xdx+ ydy)= a^2(xdy+ ydx)$
$(x^3+ xy^2)dx+ (x^2y+ y^3)dy= a^2xdy+ a^2ydx$
$(x^3+ xy^2- a^2y)dx+ (x^2y+ y^3- a^2x)dy= 0$

The derivative of $x^3+ xy^2- a^2y$ with respect to y is $2xy- a^2$ and the derivative of $x^2y+ y^3- a^2x$ with respect to x is $2xy- a^2$. Those are the same so the equation is exact.

5. ## Re: Solve first order ODE

You have made an error in the DE
To the right of the equality we have $a^2 (xdy-ydx)$

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7. ## Re: Solve first order ODE

Originally Posted by sahigg
You have made an error in the DE
To the right of the equality we have $a^2 (xdy-ydx)$
Oh, darn!

8. ## Re: Solve first order ODE

Originally Posted by sahigg
The ODE is
$xdx+ydy=\frac {a^2 (xdy-ydx)}{x^2+y^2}$

We can take $x^2+y^2$ on the left hand side to get
$(x^2+y^2)d (x^2+y^2)=2a^2 (xdy-ydx)$

I am not able to move ahead from here.
$u=x^2 + y^2$

$v = \dfrac x y$

$\dfrac{du}{dx}=2x + 2y \dfrac{dy}{dx}$

$du = 2x dx + 2y dy$

$x dx + y dy = \dfrac 1 2 du$

$\dfrac{dv}{dx} = \dfrac{y - x \frac{dy}{dx}}{y^2}$

$y^2 dv = y dx - x dy$

$\dfrac{u}{1+v^2}dv = ydx - x dy$

$a^2 \dfrac{x dy - y dx}{x^2+y^2} = -a^2 \dfrac{dv}{1+v^2}$

Now we can rewrite the original diff eq as

$\dfrac{du}{2} = -a^2 \dfrac{dv}{1+v^2}$

integrating we get

$\dfrac{u}{2} = -a^2 \arctan(v) + C$

using the original variables and rearranging we get

$\dfrac{x^2 + y^2}{2} +a^2 \arctan\left(\dfrac x y\right) = C$

and this is the implicit solution.