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

#### find_the_fun

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

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

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

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

Staff member
$$\displaystyle \sqrt{1-y^2}dx - \sqrt{1-x^2}dy=0$$, $$\displaystyle y(0)=\frac{\sqrt{3}}{2}$$

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
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
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
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
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
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
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.