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

find_the_fun

Active member
Feb 1, 2012
166
\(\displaystyle \sqrt{1-y^2}dx - \sqrt{1-x^2}dy=0\), \(\displaystyle y(0)=\frac{\sqrt{3}}{2}\)

rewriting the equation gives \(\displaystyle \frac{1}{\sqrt{1-x^2}}dx = \frac{1}{\sqrt{1-y^2}}dy\)

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

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
\(\displaystyle \sqrt{1-y^2}dx - \sqrt{1-x^2}dy=0\), \(\displaystyle y(0)=\frac{\sqrt{3}}{2}\)

rewriting the equation gives \(\displaystyle \frac{1}{\sqrt{1-x^2}}dx = \frac{1}{\sqrt{1-y^2}}dy\)

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

Well-known member
Feb 13, 2012
1,704
\(\displaystyle \sqrt{1-y^2}dx - \sqrt{1-x^2}dy=0\), \(\displaystyle y(0)=\frac{\sqrt{3}}{2}\)

rewriting the equation gives \(\displaystyle \frac{1}{\sqrt{1-x^2}}dx = \frac{1}{\sqrt{1-y^2}}dy\)

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

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \sqrt{1-y^2}dx - \sqrt{1-x^2}dy=0\), \(\displaystyle y(0)=\frac{\sqrt{3}}{2}\)

rewriting the equation gives \(\displaystyle \frac{1}{\sqrt{1-x^2}}dx = \frac{1}{\sqrt{1-y^2}}dy\)

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

Active member
Feb 1, 2012
166
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

Active member
Feb 1, 2012
166
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 \(\displaystyle C=-\frac{\pi}{3}\) because \(\displaystyle arcsin(0)=arcsin(\frac{\sqrt{3}}{2})+C\) so \(\displaystyle C=-arcsin(\frac{\sqrt{3}}{2})\)? I don't follow you here "The initial condition [FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]2[/FONT] tells you that [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]C[/FONT][FONT=MathJax_Main])"[/FONT][FONT=MathJax_Main][/FONT]
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Doesn't \(\displaystyle C=-\frac{\pi}{3}\) because \(\displaystyle arcsin(0)=arcsin(\frac{\sqrt{3}}{2})+C\) so \(\displaystyle C=-arcsin(\frac{\sqrt{3}}{2})\)? I don't follow you here "The initial condition [FONT=MathJax_Math]y[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]0[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]2[/FONT] tells you that [FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]sin[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Math]C[/FONT][FONT=MathJax_Main])"[/FONT]
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

Active member
Feb 1, 2012
166
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:

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

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

find_the_fun

Active member
Feb 1, 2012
166
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

\(\displaystyle \int \frac{1}{\sqrt{1-x^2}}dx=\int \frac{1}{\sqrt{1-y^2}}dy\) gives \(\displaystyle arcsin(x)=arcsin(y)+C\) 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

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
I get a different equation because

\(\displaystyle \int \frac{1}{\sqrt{1-x^2}}dx=\int \frac{1}{\sqrt{1-y^2}}dy\) gives \(\displaystyle arcsin(x)=arcsin(y)+C\) 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.