Diff Equs (1 Viewer)

[krypticlemonjuice]

WHere does the solution get the domain from??

 

[SB257426]

This was from the Gosford Ext 1 paper that I was doing 2 days ago so I can definetly help you out with this one !!!

In my opinion the domain is incorrect. Actually if you rearrange to make x the subject in the last line you get:


consider y=sin^-1(x)
now inverse sin of y is the inverse function of:

y= sin^-1(x)

so it will aquire a range [-1,1] because we swap the domain and the range when finding an invere function.

So the domain of our function will be [-pi/2,pi/2] so when we consider we are actually shifting it to the right by pi units because when we look at its inverse in the last line y = sin^-1(x-pi) which is a movement of pi units to the right. Thus we add pi to our extremities in the domain and the answer becomes: [pi/2,3pi/2].

 

[krypticlemonjuice]

This was from the Gosford Ext 1 paper that I was doing 2 days ago so I can definetly help you out with this one !!!

In my opinion the domain is incorrect. Actually if you rearrange to make x the subject in the last line you get:
View attachment 38964

consider y=sin^-1(x)
now inverse sin of y is the inverse function of:

y= sin^-1(x)

so it will aquire a range [-1,1] because we swap the domain and the range when finding an invere function.

So the domain of our function will be [-pi/2,pi/2] so when we consider View attachment 38964 we are actually shifting it to the right by pi units because when we look at its inverse in the last line y = sin^-1(x-pi) which is a movement of pi units to the right. Thus we add pi to our extremities in the domain and the answer becomes: [pi/2,3pi/2].

Ahhh thanks so much that makes sense

 

[howcanibesmarter]

So if you look at the original dy/dx = sqrt(1-y^2),

1-y^2 must be greater or equal to zero to be defined. Solving for y gives u the region of [-1,1]

Now plugging those values into the y=sin(x-pi) you get that domain for x, this matches SB257 answer

I hope that makes sense

 

[SB257426]

So if you look at the original dy/dx = sqrt(1-y^2),

1-y^2 must be greater or equal to zero to be defined. Solving for y gives u the region of [-1,1]

Now plugging those values into the y=sin(x-pi) you get that domain for x, this matches SB257 answer

I hope that makes sense

This is definetly a much better answer then mine cuz I went a bit too much into the detail lol. howcanibesmarter explains it in a nicer way... I would suggest you look at hers... its more concise

 

[scaryshark09]

wow, i get it, but who would even think of that in an exam