- #1

Asphyxiated

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## Homework Statement

Solve:

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

then find a solution for:

[tex] y(0)=0 [/tex]

and can you find a solution for:

[tex] y(0)=2 [/tex]

## Homework Equations

## The Attempt at a Solution

Just want to know if this is right.

First the equation can be rearranged to:

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

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

[tex] sin^{-1}(y) = x^{2}+c [/tex]

[tex] y=sin(x^{2}+c) [/tex]

for y(0)=0

[tex] 0=sin(c) [/tex]

which is valid when:

[tex] c = 0 \;\; or \;\; c=k \pi, \;\; \forall \; k \; \in \; Z [/tex]

and for y(0)=2

[tex] 2=sin(c) [/tex]

[tex] c = sin^{-1}(2) = \frac {\pi}{2}-\frac {ln(4\sqrt{3}+7)}{2} i[/tex]

which is non-real. I assume that they are looking for me to realize that the answer isn't real? I am use to dealing with complex numbers from electrical engineering so this isn't that strange to me. Although I am not sure what this really means in this context. (the context being abstract math-land.)