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Integration help?

3).

A curve passes through (0,1) and satisfies the differential equation dy/dx= (4-y^2)^1/2. Find its equation and sketch its graph.

Basically I have no idea how to integrate it?

Any hints would be appreciated.
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by Super199
3).

A curve passes through (0,1) and satisfies the differential equation dy/dx= (4-y^2)^1/2. Find its equation and sketch its graph.

Basically I have no idea how to integrate it?

Any hints would be appreciated.


use the substitution y = 2sinθ
Reply 2
Avatar for joostan
joostan
13
Original post by Super199
3).

A curve passes through (0,1) and satisfies the differential equation dy/dx= (4-y^2)^1/2. Find its equation and sketch its graph.

Basically I have no idea how to integrate it?

Any hints would be appreciated.


Use a substitution, y=2sin⁡(θ)y=2\sin(\theta) or y=2cos⁡(θ)y=2\cos(\theta) if you prefer. or look it up in a table - it's a standard result.
Reply 3
Avatar for Super199
Super199
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Original post by TeeEm
use the substitution y = 2sinθ


Original post by joostan
Use a substitution, y=2sin⁡(θ)y=2\sin(\theta) or y=2cos⁡(θ)y=2\cos(\theta) if you prefer. or look it up in a table - it's a standard result.


I think I am missing something here. What is the standard result I am supposed to see? I am not sure if were taught this, or it may be some c3/c4 stuff I have forgotten.
Reply 4
Avatar for joostan
joostan
13
Original post by Super199
I think I am missing something here. What is the standard result I am supposed to see? I am not sure if were taught this, or it may be some c3/c4 stuff I have forgotten.


The integral:
∫1a2−x2 dx\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}} \ dx
for a given constant aa is a standard result.
Reply 5
Avatar for Super199
Super199
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Original post by joostan
The integral:
∫1a2−x2 dx\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}} \ dx
for a given constant aa is a standard result.


oh right yeah that gives inverse sine right. sin^-1(x/a)+c
Reply 6
Avatar for TeeEm
TeeEm
19
Original post by Super199
I think I am missing something here. What is the standard result I am supposed to see? I am not sure if were taught this, or it may be some c3/c4 stuff I have forgotten.


It is only standard for further mathematicians,,, Just use the substitution
Reply 7
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Super199
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Original post by TeeEm
It is only standard for further mathematicians,,, Just use the substitution


It seems like a lot more effort. How does knowing the standard result help?
Reply 8
Avatar for TeeEm
TeeEm
19
Original post by Super199
It seems like a lot more effort.


I am sorry to hear that.

Original post by Super199
How does knowing the standard result help?


you can have the answer in 3 lines
(assuming you have gone the long hard substitution way a few times to have an appreciation of the techniques involved)
Reply 9
Avatar for Super199
Super199
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Original post by TeeEm
I am sorry to hear that.



you can have the answer in 3 lines
(assuming you have gone the long hard substitution way a few times to have an appreciation of the techniques involved)


Decent, care to inform me on how that is done.

Like I probably could have done it if it was 1/(4-y^2). As that is sin^-1(y/2) +c . But yeah we have one over that so how does the integral change?
Reply 10
Avatar for TeeEm
TeeEm
19
Original post by Super199
Decent, care to inform me on how that is done.

Like I probably could have done it if it was 1/(4-y^2). As that is sin^-1(y/2) +c . But yeah we have one over that so how does the integral change?


I am not following

once you separate variables you do have 1/over
Reply 11
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Super199
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Original post by TeeEm
I am not following

once you separate variables you do have 1/over


ah I got it now cheers bae
Reply 12
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TeeEm
19
Original post by Super199
ah I got it now cheers bae


all the best

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