# Thread: 9. What method I can use on this equation?

1. ## 9. What method I can use on this equation?

It is Seperation of Variable, Homogeneous Equation, Exact Equation or Linear Equation?

Question
$\displaystyle a^2(dy-dx) = x^2 dy + y^2 dx$ ; $\displaystyle a$ is constant

$\displaystyle 2$ $\displaystyle tan^{-1}$ $\displaystyle \left(\frac{y}{a}\right)$ $\displaystyle = ln \left | \frac {c(x+a)}{(x-a)} \right |$

2. Originally Posted by redfox2600
It is Seperation of Variable, Homogeneous Equation, Exact Equation or Linear Equation?

Question
$\displaystyle a^2(dy-dx) = x^2 dy + y^2 dx$ ; $\displaystyle a$ is constant

$\displaystyle 2$ $\displaystyle tan^{-1}$ $\displaystyle \left(\frac{y}{a}\right)$ $\displaystyle = ln \left | \frac {c(x+a)}{(x-a)} \right |$
It clearly re-arranges into $\displaystyle (a^2 - x^2) dy = (y^2 + a^2) dx$ and is therefore seperable.

3. Thank mr fantastic,

Here is my partial solution.

Separation of the variables:

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

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

Right side of the equation:

$\displaystyle \int \frac {dx}{(a + x)(a - x)}$

$\displaystyle \int \frac {dx}{(a + x)} \cdot \int \frac {dx}{(a-x)}$

$\displaystyle \int \frac {dx}{(x+a)} \cdot \int \frac {dx}{-(x-a)}$

$\displaystyle \int \frac {dx}{(x+a)} \cdot -\int \frac {dx}{(x-a)}$

$\displaystyle ln \mid x + a \mid$ $\displaystyle -ln \mid x - a \mid +ln \mid c \mid$

$\displaystyle ln \mid x + a \mid$ $\displaystyle - ln \mid x - a \mid +ln \mid c \mid$

$\displaystyle ln \mid x + a \mid$ $\displaystyle + ln \left( \mid x - a \mid \right)^{-1} +ln \mid c \mid$

$\displaystyle ln \left| \frac {(x + a)}{(x - a)} \right| +ln \mid c \mid$

$\displaystyle ln \left| \frac {c(x + a)}{(x - a)} \right|$

Did I wrote the equation correctly? Correct me if I am wrong.

Left side of the equation:

$\displaystyle \int \frac {dy}{(a^2 + y^2)}$
I'm stucked in this point. How can I go to the answer (left side of the equation)?

$\displaystyle 2 tan^{-1} \frac {y}{a} = ln \left| \frac {c(x+a)}{(x-a)} \right|$

I think this theorem would be help. But I am not sure how to go to the answer.
Theorem: Inverse Trigonometric Functions
$\displaystyle \frac {d}{dx} (tan^{-1}) = \frac {1}{1 + u^2} \frac {du}{dx}$

Thanks to all.
Regards,

4. Originally Posted by redfox2600
$\displaystyle \int \frac {dx}{(x+a)} \cdot -\int \frac {dx}{(x-a)}$
$\displaystyle ln \mid x + a \mid$ $\displaystyle -ln \mid x - a \mid +ln \mid c \mid$
This is wrong. It's multiplication, not substraction
Use partial fraction

Left side of the equation:

$\displaystyle \int \frac {dy}{(a^2 + y^2)}$
I'm stucked in this point. How can I go to the answer (left side of the equation)?
use substitution : y = a (tan theta)

5. thanks songoku.