solve the ivp dy/dx = 1/(sqrt(1-x^2)) with initial condition y(sqrt(3)/2)=0

1 answer

y = arcsin(x) + c

0 = arcsin(√3/2) + c
0 = π/3 + c
c = -π/3

so,
y = arcsin(x) - π/3
Similar Questions
  1. When I solve the inquality 2x^2 - 6 < 0,I get x < + or - sqrt(3) So how do I write the solution? Is it (+sqrt(3),-sqrt(3)) or
    1. answers icon 0 answers
  2. Please look at my work below:Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/-
    1. answers icon 0 answers
  3. Evaluate sqrt7x (sqrt x-7 sqrt7) Show your work.sqrt(7)*sqrt(x)-sqrt(7)*7*sqrt(7) sqrt(7*x)-7*sqrt(7*7) sqrt(7x)-7*sqrt(7^2)
    1. answers icon 1 answer
    1. answers icon 2 answers
more similar questions