Solving Initial Value Problem: dy/dx = (1-y^2)tanx?

Question

How do I solve this initial value problem?

dy/dx = ( 1 + (y^2)) tanx

y(0)=3^0.5

I separated the variables and got:

tan^-1 (y) = -ln(|cos x|)

But that doesn't make sense...

Please help!

Answer 1

dy/dx = (1-y²) tanx

dy/(1-y²) = tanx dx

the integral of 1/(1-y²) is NOT arctan(y)

it's 1/(1+y²) whose integral is arctan(y) --- Note: that is a PLUS sign, not negative sign. Your equation has the negative sign in it

recall that ∫du/(a²-u²) = 1/(2a) Ln l (a+u)/(a-u) l + C

so on the left side, you should have had:

1/2 Ln l (1+y)/(1-y) l = ln l secx l + C

Note: you forgot to add C, a constant

and -ln l cosx l = ln l secx l

they gave you y(0) = 3^.5

1/2 Ln l (1+3^.5)/(1-3^.5) l = ln l sec0 l + C

C =~ 0.65847

so you have:

1/2 Ln l (1+y)/(1-y) l = ln l secx l + 0.65847

Now just do the algebra to solve for y. Whatever you get y equals, that's your solution