MHB Getting wrong answer to differential equation (first order separable ODE)

[find_the_fun]

[math]\sqrt{1-y^2}dx - \sqrt{1-x^2}dy=0[/math], [math]y(0)=\frac{\sqrt{3}}{2}[/math]

rewriting the equation gives [math]\frac{1}{\sqrt{1-x^2}}dx = \frac{1}{\sqrt{1-y^2}}dy[/math]

Isn't this the integral for [math]\sin^{-1}(x)[/math] & [math]\sin^{-1}(y)[/math]? The back of book has [math]y=1/2x+\frac{\sqrt{3}}{2}\sqrt{1-x^2}[/math]

 

[Chris L T521]

[math]\sqrt{1-y^2}dx - \sqrt{1-x^2}dy=0[/math], [math]y(0)=\frac{\sqrt{3}}{2}[/math]

rewriting the equation gives [math]\frac{1}{\sqrt{1-x^2}}dx = \frac{1}{\sqrt{1-y^2}}dy[/math]

Isn't this the integral for [math]\sin^{-1}(x)[/math] & [math]\sin^{-1}(y)[/math]? The back of book has [math]y=1/2x+\frac{\sqrt{3}}{2}\sqrt{1-x^2}[/math]

Well, assuming that is indeed the correct problem, then your set up is correct. Solving the ODE gives you $y=\sin\left(\arcsin (x) + C\right)$.

The initial condition $y(0)=\dfrac{\sqrt{3}}{2}$ tells you that $\dfrac{\sqrt{3}}{2}=\sin\left(C\right) \implies C= \dfrac{\pi}{3}+2k\pi,\dfrac{2\pi}{3}+2k\pi$. Since we can't have an infinite number of constants, take $C$ to be the principal value, i.e. $C=\dfrac{\pi}{3}$,

Then $y(x) = \sin\left(\arcsin x + \dfrac{\pi}{3}\right)$.

Now to get the answer they have, apply the identity $\sin(\alpha+\beta) = \sin\alpha\cos\beta+\cos\alpha\sin\beta$ and simplify.

I hope this makes sense!

 

[chisigma]

[math]\sqrt{1-y^2}dx - \sqrt{1-x^2}dy=0[/math], [math]y(0)=\frac{\sqrt{3}}{2}[/math]

rewriting the equation gives [math]\frac{1}{\sqrt{1-x^2}}dx = \frac{1}{\sqrt{1-y^2}}dy[/math]

Isn't this the integral for [math]\sin^{-1}(x)[/math] & [math]\sin^{-1}(y)[/math]? The back of book has [math]y=1/2x+\frac{\sqrt{3}}{2}\sqrt{1-x^2}[/math]

The general solution You arrive to is...

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

... and from (1) You derive...

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

Now You use the initial value and find c...

Kind regards

$\chi$ $\sigma$

 

[MarkFL]

[math]\sqrt{1-y^2}dx - \sqrt{1-x^2}dy=0[/math], [math]y(0)=\frac{\sqrt{3}}{2}[/math]

rewriting the equation gives [math]\frac{1}{\sqrt{1-x^2}}dx = \frac{1}{\sqrt{1-y^2}}dy[/math]

Isn't this the integral for [math]\sin^{-1}(x)[/math] & [math]\sin^{-1}(y)[/math]? The back of book has [math]y=1/2x+\frac{\sqrt{3}}{2}\sqrt{1-x^2}[/math]

I edited the thread title and the body of the post to rewrite the n's in lowercase and added a bit of information to the title to describe the problem.

 

[find_the_fun]

I edited the thread title and the body of the post to rewrite the n's in lowercase and added a bit of information to the title to describe the problem.

Yea sorry about that my keyboard is breaking down.

 

[find_the_fun]

Well, assuming that is indeed the correct problem, then your set up is correct. Solving the ODE gives you $y=\sin\left(\arcsin (x) + C\right)$.

The initial condition $y(0)=\dfrac{\sqrt{3}}{2}$ tells you that $\dfrac{\sqrt{3}}{2}=\sin\left(C\right) \implies C= \dfrac{\pi}{3}+2k\pi,\dfrac{2\pi}{3}+2k\pi$. Since we can't have an infinite number of constants, take $C$ to be the principal value, i.e. $C=\dfrac{\pi}{3}$,

Then $y(x) = \sin\left(\arcsin x + \dfrac{\pi}{3}\right)$.

Now to get the answer they have, apply the identity $\sin(\alpha+\beta) = \sin\alpha\cos\beta+\cos\alpha\sin\beta$ and simplify.

I hope this makes sense!

Doesn't [math]C=-\frac{\pi}{3}[/math] because [math]arcsin(0)=arcsin(\frac{\sqrt{3}}{2})+C[/math] so [math]C=-arcsin(\frac{\sqrt{3}}{2})[/math]? I don't follow you here "The initial condition [FONT=MathJax_Math]y[FONT=MathJax_Main]([FONT=MathJax_Main]0[FONT=MathJax_Main])[FONT=MathJax_Main]=[FONT=MathJax_Main]3[FONT=MathJax_Main]√[FONT=MathJax_Main]2 tells you that [FONT=MathJax_Main]3[FONT=MathJax_Main]√[FONT=MathJax_Main]2[FONT=MathJax_Main]=[FONT=MathJax_Main]sin[FONT=MathJax_Main]([FONT=MathJax_Math]C[FONT=MathJax_Main])"[FONT=MathJax_Main]

 

[Chris L T521]

Doesn't [math]C=-\frac{\pi}{3}[/math] because [math]arcsin(0)=arcsin(\frac{\sqrt{3}}{2})+C[/math] so [math]C=-arcsin(\frac{\sqrt{3}}{2})[/math]? I don't follow you here "The initial condition [FONT=MathJax_Math]y[FONT=MathJax_Main]([FONT=MathJax_Main]0[FONT=MathJax_Main])[FONT=MathJax_Main]=[FONT=MathJax_Main]3[FONT=MathJax_Main]√[FONT=MathJax_Main]2 tells you that [FONT=MathJax_Main]3[FONT=MathJax_Main]√[FONT=MathJax_Main]2[FONT=MathJax_Main]=[FONT=MathJax_Main]sin[FONT=MathJax_Main]([FONT=MathJax_Math]C[FONT=MathJax_Main])"

If $y(x) = \sin\left(\arcsin x + C\right)$, then $y(0) = \dfrac{\sqrt{3}}{2} \implies \dfrac{\sqrt{3}}{2} = \sin(\arcsin 0 + C) = \sin C$.

Thus $\dfrac{\sqrt{3}}{2}=\sin C \implies C=\arcsin\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{3}$.

If you want to leave things in terms of inverse sines, then the initial condition $y(0)=\dfrac{\sqrt{3}}{2}$ would the get plugged in as follows:

\[\arcsin y = \arcsin x + C \xrightarrow{y(0)=\dfrac{\sqrt{3}}{2}}{} \arcsin\left(\frac{\sqrt{3}}{2}\right) = \arcsin 0 + C \implies C = \arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}.\]

I hope this clarifies things!

 

[find_the_fun]

If $y(x) = \sin\left(\arcsin x + C\right)$, then $y(0) = \dfrac{\sqrt{3}}{2} \implies \dfrac{\sqrt{3}}{2} = \sin(\arcsin 0 + C) = \sin C$.

Thus $\dfrac{\sqrt{3}}{2}=\sin C \implies C=\arcsin\left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{3}$.

If you want to leave things in terms of inverse sines, then the initial condition $y(0)=\dfrac{\sqrt{3}}{2}$ would the get plugged in as follows:

\[\arcsin y = \arcsin x + C \xrightarrow{y(0)=\dfrac{\sqrt{3}}{2}}{} \arcsin\left(\frac{\sqrt{3}}{2}\right) = \arcsin 0 + C \implies C = \arcsin\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}.\]

I hope this clarifies things!

I see your way but is there something wrong with the following approach:

[math]\int \frac{1}{\sqrt{1-x^2}} dx = \int \frac{1}{\sqrt{1-y^2}} dy[/math] gives [math]arcsin(x)=arcsin(y)+C[/math] using the initial condition
[math]arcsin(0)=arcsin(\frac{\sqrt{3}}{2})+C[/math]
[math]0=a\frac{\pi}{3}+C[/math]
therefore [math]C=-\frac{\pi}{-3}[/math] note the negative sign.

Oh I see I switched what I should be plugging in for x and y.

 

[find_the_fun]

If you want to leave things in terms of inverse sines, then the initial condition $y(0)=\dfrac{\sqrt{3}}{2}$ would the get plugged in as follows:

\[\arcsin y = \arcsin x + C\]...

I get a different equation because

[math]\int \frac{1}{\sqrt{1-x^2}}dx=\int \frac{1}{\sqrt{1-y^2}}dy[/math] gives [math]arcsin(x)=arcsin(y)+C[/math] How come your C is on the other side of the equation? Do you get to decide which side of the equation has the arbitrary constant? Doesn't it change the sign?

 

[Prove It]

I get a different equation because

[math]\int \frac{1}{\sqrt{1-x^2}}dx=\int \frac{1}{\sqrt{1-y^2}}dy[/math] gives [math]arcsin(x)=arcsin(y)+C[/math] How come your C is on the other side of the equation? Do you get to decide which side of the equation has the arbitrary constant? Doesn't it change the sign?

Your arbitrary constant is just that, ARBITRARY. You can write it in any form you like. When you solve for it, you'll get the same answer.