How would you solve the following problem explicitly?
Sqrt(1-y^2) dx - sqrt(1-x^2) dy
I separated the x and y terms and got:
Integral of 1/sqrt(1-x^2) dx = Integral of 1/sqrt(1-y^2) dy
I was wondering how would you take the anti-derivative of each function. I believe we are using a trig substitution?
3 answers
Oops, Sqrt(1-y^2) dx - sqrt(1-x^2) dy = 0
sqrt(1-y^2) dx - sqrt(1-x^2) dy = 0
sqrt(1-y^2) dx = sqrt(1-x^2) dy
Take all term swith x to one side and all terms of y to the other side of equation:
dx / sqrt(1-x^2) = dy / sqrt(1-y^2)
Integrate both side. Note that,
∫(1/(sqrt(a^2 - x^2))dx = sin^-1 (x/a) + C
Therefore,
sin^-1 (x) + C = sin^-1 (y)
Get the sine of both sides:
y = sin( sin^-1 (x) + C )
hope this helps? `u`
sqrt(1-y^2) dx = sqrt(1-x^2) dy
Take all term swith x to one side and all terms of y to the other side of equation:
dx / sqrt(1-x^2) = dy / sqrt(1-y^2)
Integrate both side. Note that,
∫(1/(sqrt(a^2 - x^2))dx = sin^-1 (x/a) + C
Therefore,
sin^-1 (x) + C = sin^-1 (y)
Get the sine of both sides:
y = sin( sin^-1 (x) + C )
hope this helps? `u`
If we pick a different c, the solution looks a little more symmetric, and more like the original equation:
y = sin(arcsin(x)+arcsin(c))
y = sin(arcsin(x))cos(arcsin(c)) + cos(arcsin(x))sin(arcsin(c))
= x√(1-c^2) + c√(1-x^2)
Just a thought. A bit ugly because c appears twice, and is more restricted than the general C above.
y = sin(arcsin(x)+arcsin(c))
y = sin(arcsin(x))cos(arcsin(c)) + cos(arcsin(x))sin(arcsin(c))
= x√(1-c^2) + c√(1-x^2)
Just a thought. A bit ugly because c appears twice, and is more restricted than the general C above.