Question

Solve this DE:

Answer 1

\[\sqrt{1-y^2}dx-\sqrt{1-x^2}dy=0\]

Answer 2

\[\sqrt{1-y^2}dx=\sqrt{1-x^2}dy\]

Answer 3

\[\frac{dx}{\sqrt{1-x^2}}=\frac{dy}{\sqrt{1-y^2}}\]

Answer 4

We can write this as
\[\frac{dy}{\sqrt{1-y^2}}=\frac{dx}{\sqrt{1-x^2}}\]
Now integrate both sides
\[\int\frac{dy}{\sqrt{1-y^2}}=\int \frac{dx}{\sqrt{1-x^2}}\]
This is a standard integral, can you do this ?

Answer 5

yeah it's arcsin however it asks what specific solution at y(square root of 3 over 2)

arcsin(y)=arcsin(x)+c


should i multiply both sides by sin to cancel and get in y or should i just keep this like so

Answer 6

y=x+sin(c)

Answer 7

i guess i should keep since i'd have to take the arcsin to get back to c

Answer 8

Is the value of x given?

Answer 9

yeah y(0)= sqrt 3 / 2

Answer 10

put x=0
\[\sin^{-1} \frac{\sqrt 3}{2}=\sin^{-1}(0)+c\]
\[\frac{\pi}{3}=0+c\]
Now our DE is
\[\sin^{-1} y=\sin^{-1}x+\frac{\pi}{3}\]