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FP2 Differential equations Help

Find the solution to the differential equation (4-x^2)^1/2 dy/dx = (4-y^2) ^1/2 for which y= -1 and x=1.

I just don't know how to integrate (4-y^2)^1/2 = (4-y^2)^1/2

Can someone please help
Original post by ChemBoy1
Find the solution to the differential equation (4-x^2)^1/2 dy/dx = (4-y^2) ^1/2 for which y= -1 and x=1.

I just don't know how to integrate (4-y^2)^1/2 = (4-y^2)^1/2

Can someone please help


Not sure what the second line means.

Use the substitution x=2sinu.
Original post by ChemBoy1
Find the solution to the differential equation (4-x^2)^1/2 dy/dx = (4-y^2) ^1/2 for which y= -1 and x=1.

I just don't know how to integrate (4-y^2)^1/2 = (4-y^2)^1/2

Can someone please help


(4−x2)1/2dydx=(4−y2)1/2⇔(4−y2)−1/2.dy=(4−x2)−1/2.dx⇔∫(4−y2)−1/2.dy=∫(4−x2)−1/2.dx\begin{aligned} (4-x^2)^{1/2} \dfrac{dy}{dx} = (4-y^2)^{1/2} & \Leftrightarrow (4-y^2)^{-1/2} .dy = (4-x^2)^{-1/2} .dx \\ & \Leftrightarrow \int (4-y^2)^{-1/2} .dy = \int (4-x^2)^{-1/2} .dx \end{aligned}

Then just use two subs:

x=2sin⁡ux= 2 \sin u
y=2sin⁡vy= 2 \sin v
Update: I managed to get the equation as arcsin y/2 = arcsin x/2 +c. The question asks to express my answer in a form not involving trig functions. I don't see how I can do that.

Thanks
Original post by ChemBoy1
Update: I managed to get the equation as arcsin y/2 = arcsin x/2 +c. The question asks to express my answer in a form not involving trig functions. I don't see how I can do that.

Thanks


First of all, determine what cc is.

Then, write y2=sin⁡(arcsin⁡x2+c)\dfrac{y}{2}=\sin \left( \arcsin \dfrac{x}{2} + c \right) and then apply the compound angle formula on the RHS.
I got c as pi/3. Ho do I convert it to an expression not involving trig functions?
Original post by RDKGames
First of all, determine what cc is.

Then, write y2=sin⁡(arcsin⁡x2+c)\dfrac{y}{2}=\sin \left( \arcsin \dfrac{x}{2} + c \right) and then apply the compound angle formula on the RHS.
Original post by ChemBoy1
I got c as pi/3. Ho do I convert it to an expression not involving trig functions?


Read my post - I answered that question...
Sorry just saw that thanks. How can you use compound angle formula because i still got Sin (Sin-1 x) etc.
Original post by RDKGames
Read my post - I answered that question...
Original post by ChemBoy1
Sorry just saw that thanks. How can you use compound angle formula because i still got Sin (Sin-1 x) etc.


sin⁡arcsin⁡x2=x2\sin \arcsin \frac{x}{2} = \frac{x}{2} (obviously) so, we just need cos⁡arcsin⁡x2\cos \arcsin \frac{x}{2}. If we denote arcsin⁡x2=θ\arcsin \frac{x}{2} = \theta then we have sin⁡θ=x2\sin \theta = \frac{x}{2} and we can use a right-angled triangle to determine what cos⁡θ\cos \theta is.

EDIT: Or just use cos⁡θ=1−sin⁡2θ\cos \theta = \sqrt{1-\sin^2 \theta} for this part.
(edited 6 years ago)
Still confused with the right angle triangle thing. becuase the hyp = 2 and the opp = x. The adj would just be a separate variable and that would not work? Could you please help me out?
Original post by RDKGames
sin⁡arcsin⁡x2=x2\sin \arcsin \frac{x}{2} = \frac{x}{2} (obviously) so, we just need cos⁡arcsin⁡x2\cos \arcsin \frac{x}{2}. If we denote arcsin⁡x2=θ\arcsin \frac{x}{2} = \theta then we have sin⁡θ=x2\sin \theta = \frac{x}{2} and we can use a right-angled triangle to determine what cos⁡θ\cos \theta is.

EDIT: Or just use cos⁡θ=1−sin⁡2θ\cos \theta = \sqrt{1-\sin^2 \theta} for this part.
Original post by ChemBoy1
Still confused with the right angle triangle thing. becuase the hyp = 2 and the opp = x. The adj would just be a separate variable and that would not work? Could you please help me out?


The adj would just be sqrt(hyp^2-opp^2) by Pythagoras
The book got y= 1/2 x - 1/2x (3(4-x^2))^1/2.

I got no where near that answer, could you help elaborate.
FYI values are y= -1 and x=1
Original post by ChemBoy1
The book got y= 1/2 x - 1/2x (3(4-x^2))^1/2.

I got no where near that answer, could you help elaborate.
FYI values are y= -1 and x=1


OK, first of all, c≠π3c \neq \frac{\pi}{3}. It is in fact c=−π3c= -\frac{\pi}{3}

Next, we have
y2=sin⁡(arcsin⁡x2−π3)=sin⁡(arcsin⁡x2)cos⁡(π3)−sin⁡(π3)cos⁡(arcsin⁡x2)=x2⋅12−32⋅1−x24\begin{aligned} \dfrac{y}{2} & = \sin \left( \arcsin \frac{x}{2} - \frac{\pi}{3} \right) \\ & = \sin \left( \arcsin \frac{x}{2} \right) \cos \left( \frac{\pi}{3} \right) - \sin \left( \frac{\pi}{3} \right) \cos \left( \arcsin \frac{x}{2} \right) \\ & = \frac{x}{2} \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^2}{4}} \end{aligned}

Finish it off.
Nearly got it, except for how they got 3(4-x^2)^1/2
Original post by RDKGames
OK, first of all, c≠π3c \neq \frac{\pi}{3}. It is in fact c=−π3c= -\frac{\pi}{3}

Next, we have
y2=sin⁡(arcsin⁡x2−π3)=sin⁡(arcsin⁡x2)cos⁡(π3)−sin⁡(π3)cos⁡(arcsin⁡x2)=x2⋅12−32⋅1−x24\begin{aligned} \dfrac{y}{2} & = \sin \left( \arcsin \frac{x}{2} - \frac{\pi}{3} \right) \\ & = \sin \left( \arcsin \frac{x}{2} \right) \cos \left( \frac{\pi}{3} \right) - \sin \left( \frac{\pi}{3} \right) \cos \left( \arcsin \frac{x}{2} \right) \\ & = \frac{x}{2} \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^2}{4}} \end{aligned}

Finish it off.
Original post by ChemBoy1
Nearly got it, except for how they got 3(4-x^2)^1/2


Factor out 1/4 inside the root, then bring that factor outside the root. Similarly, merge root(3) with the main root since ab=ab\sqrt{a}\sqrt{b} = \sqrt{ab}
Thank you so much! You have helped a lot! Sorry if I have sounded very silly to you.
Original post by RDKGames
Factor out 1/4 inside the root, then bring that factor outside the root. Similarly, merge root(3) with the main root since ab=ab\sqrt{a}\sqrt{b} = \sqrt{ab}
Original post by ChemBoy1
Thank you so much! You have helped a lot! Sorry if I have sounded very silly to you.


:thumbsup:

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