Differential Equation: Initial Value

**twoteenine** · March 15th 2010, 11:43 AM

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

I changed the equation to get

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

Took the integral to get:

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

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

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

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

**Bruno J.** · March 15th 2010, 11:51 AM

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

**harish21** · March 15th 2010, 01:35 PM

Originally Posted by twoteenine

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

I changed the equation to get

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

Took the integral to get:

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

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

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

When 0 is plugged in for x though, $\frac{1}{sin^{-1}(0)}$ 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

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

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

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

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

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

**skeeter** · March 15th 2010, 01:56 PM

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

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

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

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

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

Bruno J's suggestion follows ...

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

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

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

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

...

**harish21** · March 15th 2010, 02:09 PM

Originally Posted by skeeter

...

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

Thread: Differential Equation: Initial Value

LinkBack

Thread Tools

Display

Differential Equation: Initial Value

Similar Math Help Forum Discussions

Differential equation with initial conditions

Differential equation and initial condition

Initial Value Problem - Differential Equation!!!

[SOLVED] initial value differential equation

differential equation w/ initial condition

Search Tags