# Thread: Differential Equation: Initial Value

1. ## Differential Equation: Initial Value

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

I changed the equation to get

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

Took the integral to get:

$\displaystyle -y^{-1} = sin^{-1}(x) + C$

I'm not sure what the rules with $\displaystyle sin^{-1}(x)$ are, but i got y by itself, ending with

$\displaystyle y = -\frac{1}{sin^{-1}(x)} + C$

When 0 is plugged in for x though, $\displaystyle \frac{1}{sin^{-1}(0)}$ goes to infinity. What did i do wrong?

2. Be careful with your integration constant! It should be on the denominator once you've solved for $\displaystyle y$.

3. Originally Posted by twoteenine
$\displaystyle \frac{dy}{dx} \sqrt{1-x^{2}} = y^{2} , y(0) = 1$

I changed the equation to get

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

Took the integral to get:

$\displaystyle -y^{-1} = sin^{-1}(x) + C$

I'm not sure what the rules with $\displaystyle sin^{-1}(x)$ are, but i got y by itself, ending with

$\displaystyle y = -\frac{1}{sin^{-1}(x)} + C$

When 0 is plugged in for x though, $\displaystyle \frac{1}{sin^{-1}(0)}$ goes to infinity. What did i do wrong?
Apart from your solution going to infinity, there is an error in your differentiation on the LHS of the equation you have stated

$\displaystyle \frac{dy}{y^2} = \frac{dx}{\sqrt{1-x^2}}$ gives you

$\displaystyle -2y^{-1} = sin^{-1}(x) + C$

or, $\displaystyle - sin^{-1}(x) + C'= \frac{2}{y}$

or, $\displaystyle y = \frac{2}{- sin^{-1}(x) + C'}$

substitute y(0) = 1 to get $\displaystyle C' = 2$

4. Originally Posted by harish21
Apart from your solution going to infinity, there is an error in your differentiation on the LHS of the equation you have stated

$\displaystyle \frac{dy}{y^2} = \frac{dx}{\sqrt{1-x^2}}$ gives you

$\displaystyle -\textcolor{red}{2}y^{-1} = sin^{-1}(x) + C$

no 2 ... twoteenine's original antideruivative is correct.

$\displaystyle \textcolor{red}{-\frac{1}{y} = \sin^{-1}(x) + C}$

Bruno J's suggestion follows ...

$\displaystyle \textcolor{red}{y = -\frac{1}{\sin^{-1}(x) + C}}$

$\displaystyle \textcolor{red}{y(0) = 1}$

$\displaystyle \textcolor{red}{C = -1}$

$\displaystyle \textcolor{red}{y = \frac{1}{1 - \sin^{-1}(x)}}$
...

5. Originally Posted by skeeter
...
Thanks for the correction skeeter. I did not integrate the LHS. Instead I was doing something stupid there!