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General solution of first order ODE stuck on integrating

I'm stuck on the following differential equation question, I have to solve it and find a general solution in explicit form:
dydx=(1y2)×sin(x)\displaystyle\frac{dy}{dx} = \sqrt{(1-y^2)} \times sin(x)

My attempt is as follows, using the variable separable method:
1(1y2)12dydx=sin(x)\displaystyle\frac{1}{(1-y^2)^\frac{1}{2}} \frac{dy}{dx} = sin(x)

1(1y2)12dy=sin(x)dx\displaystyle\int \frac{1}{(1-y^2)^\frac{1}{2}} dy = \int sin(x) dx

Here is where I got stuck integrating the LHS. Am I correct in integrating by substitution with y = sinx?

I tried it but the answer I got looked wrong, I got upto (for LHS):
sec2(x)dy\displaystyle\int sec^2(x) dy

I'm not sure whether my integration by trig substitution is the problem or if this type of ODE doesn't work for the variable separation method, any insights would be appreciated.

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For the substitution, you will need to introduce another variable, so y=sinuy = \sin u is OK. Then recall that ddutanu=sec2u\frac{d}{du}\tan u = \sec^2 u.
Reply 2
Original post by Sayless
I'm stuck on the following differential equation question, I have to solve it and find a general solution in explicit form:
dydx=(1y2)×sin(x)\displaystyle\frac{dy}{dx} = \sqrt{(1-y^2)} \times sin(x)

My attempt is as follows, using the variable separable method:
1(1y2)12dydx=sin(x)\displaystyle\frac{1}{(1-y^2)^\frac{1}{2}} \frac{dy}{dx} = sin(x)

1(1y2)12dy=sin(x)dx\displaystyle\int \frac{1}{(1-y^2)^\frac{1}{2}} dy = \int sin(x) dx

Here is where I got stuck integrating the LHS. Am I correct in integrating by substitution with y = sinx?

I tried it but the answer I got looked wrong, I got upto (for LHS):
sec2(x)dy\displaystyle\int sec^2(x) dy

I'm not sure whether my integration by trig substitution is the problem or if this type of ODE doesn't work for the variable separation method, any insights would be appreciated.



That is the correct substitution, however it seems like you forgot to square root the denominator after the sub

Also remember that if y=sinx then dy/dx = cosx and so dy=cosx dx meaning the numerator and denominator should end up cancelling out

Or just use the formula book since it's a standard integral
Reply 3
Original post by A Slice of Pi
For the substitution, you will need to introduce another variable, so y=sinuy = \sin u is OK. Then recall that ddutanu=sec2u\frac{d}{du}\tan u = \sec^2 u.


with sec2(x)dy\displaystyle\int sec^2(x) dy
I have to integrate with respect to y, and I have an x value so I imagined it as a constant and having a 1 infront of it so turning that into a y with the constant following, I'm not sure what to do here.
I can see what you mean and I would do that if the integral was with respect to x, do you end up with my answer when you try the substitution method?
Reply 4
Photo 04-02-2017 18 51 07.jpg
I've attached my working here, did I substitute it right?
Reply 5
Original post by Sayless

I've attached my working here, did I substitute it right?


No, this is what happens when you get too used to what "x" is meant to mean.

You want to substitute for dy in terms of dx, because your original integral is with respect to dy and you want to change it to dx. So dy = cos x dx.

Then 1cosxdy=1cosxcosxdx=1dx\int \frac{1}{\cos x} \, \mathrm{d}y = \int \frac{1}{\cos x} \cdot \cos x \, \mathrm{d}x = \int 1 \, \mathrm{d}x

If I were you, I'd just avoid using x as a substitution variable and stick with u or something...
Reply 6
Original post by Zacken
No, this is what happens when you get too used to what "x" is meant to mean.

You want to substitute for dy in terms of dx, because your original integral is with respect to dy and you want to change it to dx. So dy = cos x dx.

Then 1cosxdy=1cosxcosxdx=1dx\int \frac{1}{\cos x} \, \mathrm{d}y = \int \frac{1}{\cos x} \cdot \cos x \, \mathrm{d}x = \int 1 \, \mathrm{d}x

If I were you, I'd just avoid using x as a substitution variable and stick with u or something...


Ah, I see where I went wrong now thanks for pointing it out. You're right I'm too used to subbing the dy into the dx so I did it out of habit without thinking about it, I'll begin using u instead
Original post by Sayless
Photo 04-02-2017 18 51 07.jpg
I've attached my working here, did I substitute it right?


use the fact that 11y2dy=arcsin(y)+c\int\frac{1}{\sqrt{1-y^{2}}}\text{d}y=\text{arcsin}(y)+c
Reply 8
Original post by Cryptokyo
use the fact that 11y2dy=arcsin(y)+c\int\frac{1}{\sqrt{1-y^{2}}}\text{d}y=\text{arcsin}(y)+c


Photo 04-02-2017 19 15 57.jpg
I'm trying that now and I got the general solution in explicit form but I'm having trouble verifying it, I've attached my working.
I need the sqrt(1-y^2) to be sin(C-cosx) to verify it but I can only get it in terms of cos, am I doing it right?

if my working is unclear i can rewrite it clearly including full question, it's all scattered atm, the question is in the first post
(edited 7 years ago)
Original post by Sayless
Photo 04-02-2017 19 15 57.jpg
I'm trying that now and I got the general solution in explicit form but I'm having trouble verifying it, I've attached my working.
I need the sqrt(1-y^2) to be sin(C-cosx) to verify it but I can only get it in terms of cos, am I doing it right?

if my working is unclear i can rewrite it clearly including full question, it's all scattered atm, the question is in the first post


Sorry for the late reply.

You differentiation is incorrect. The derivative of sin(Ccos(x))\sin(C-\cos(x)) is sinx×cos(Ccos(x))\sin x \times \cos(C-\cos(x)) and not sinx×sin(Ccos(x))\sin x \times \sin(C-\cos (x)) It should work then.
(edited 7 years ago)
Reply 10
Original post by Cryptokyo
Sorry for the late reply.

You differentiation is incorrect. The derivative of sin(Ccos(x))\sin(C-\cos(x)) is sinx×cos(Ccos(x))\sin x \times \cos(C-\cos(x)) and not sinx×sin(Ccos(x))\sin x \times \sin(C-\cos (x)) It should work then.


Ah I've got it now thank you,
1/a ln(ax+b) is the solution for integration a fraction. When a is 1 and the numerator is one then the integral is just ln|f(x)|
Original post by Anfanny
1/a ln(ax+b) is the solution for integration a fraction. When a is 1 and the numerator is one then the integral is just ln|f(x)|
It's the general solution for integrating 1/(ax+b).

The integral here is not of that form.
Reply 13
Original post by Cryptokyo
use the fact that 11y2dy=arcsin(y)+c\int\frac{1}{\sqrt{1-y^{2}}}\text{d}y=\text{arcsin}(y)+c


And why's that true...?
@Sayless You may be interested in this...

Original post by Zacken
And why's that true...?


Let y=arcsin(x)y=\text{arcsin}(x)
x=siny\Rightarrow x=\sin y
dxdy=cosy\frac{dx}{dy}=\cos y
Using dydxdxdy=1\frac{dy}{dx}\frac{dx}{dy}=1
dydx=1cosy=11sin2y=11x2\frac{dy}{dx}= \frac{1}{\cos y}= \frac{1}{\sqrt{1-\sin^{2}y}}= \frac{1}{\sqrt{1-x^{2}}}
Original post by Cryptokyo
@Sayless You may be interested in this...



Let y=arcsin(x)y=\text{arcsin}(x)
x=siny\Rightarrow x=\sin y
dxdy=cosy\frac{dx}{dy}=\cos y
Using dydxdxdy=1\frac{dy}{dx}\frac{dx}{dy}=1
dydx=1cosy=11sin2y=11x2\frac{dy}{dx}= \frac{1}{\cos y}= \frac{1}{\sqrt{1-\sin^{2}y}}= \frac{1}{\sqrt{1-x^{2}}}
Technically, this is correct. But (I'm sure) the point Zacken is trying to make here is that you should be able to find this result by substitution.

To be clear, instead of using substitution, you can do this by recognition (which is kind of what you've done, otherwise you'd have no reason to "start" with arcsin x), or by looking it up in the formula book, and these are all fine valid methods.

But if someone tries to do this by substitution, and they can't, then they need to brush up on your substitution methods; someone telling them "it's easy to verify that arcsin x works" is not doing them a service in the long term.
Original post by Sayless
with sec2(x)dy\displaystyle\int sec^2(x) dy
I have to integrate with respect to y, and I have an x value so I imagined it as a constant and having a 1 infront of it so turning that into a y with the constant following, I'm not sure what to do here.
I can see what you mean and I would do that if the integral was with respect to x, do you end up with my answer when you try the substitution method?

You have made the substitution y=sinx y = \sin x and your intuition is correct. However, you will need to use another variable not aready present, say uu, instead of xx. Remember the goal here. You're solving the ODE to find yy as a function of xx. When you write y=sinx y = \sin x you're actually claiming that this is the solution of the ODE!
Reply 17
Original post by A Slice of Pi
You have made the substitution y=sinx y = \sin x and your intuition is correct. However, you will need to use another variable not aready present, say uu, instead of xx. Remember the goal here. You're solving the ODE to find yy as a function of xx. When you write y=sinx y = \sin x you're actually claiming that this is the solution of the ODE!


Ah I used x because the RHS was in terms of x so I thought it would be easier, it still works with x but for better presentation I should use u instead?
Original post by Sayless
Ah I used x because the RHS was in terms of x so I thought it would be easier, it still works with x but for better presentation I should use u instead?

No, we use uu instead of xx because xx is already present on the RHS. Never use xx for substitutions if it is already present somewhere else!
Reply 19
Original post by DFranklin
Technically, this is correct. But (I'm sure) the point Zacken is trying to make here is that you should be able to find this result by substitution.


PRSOM

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