Find the particular solution to the differential equation y√ (1-x^2) dy/dx - x√ (1-y^2)=0 given the initial condition y(0)=1
To find the particular solution to the differential equation, we first need to rearrange it into a form that we can integrate.
Given differential equation: y√(1-x^2)dy/dx - x√(1-y^2) = 0
Multiplying through by dx, we get:
y√(1-x^2)dy = x√(1-y^2)dx
Next, we can separate variables by dividing both sides by y√(1-y^2) and multiplying both sides by dx:
dy/(y√(1-y^2)) = xdx/(√(1-x^2))
Integrating both sides:
∫(1/(y√(1-y^2))) dy = ∫(x/(√(1-x^2))) dx
Let u = y^2, then du = 2y dy
Let v = x^2, then dv = 2x dx
Substitute u and v:
1/2 ∫(1/(√(1-u))) du = 1/2 ∫(1/(√(1-v))) dv
Integrating both sides:
1/2 arcsin(u) + C1 = 1/2 arcsin(v) + C2
arcsin(u) = arcsin(v) + C
Now substitute back u = y^2 and v = x^2:
arcsin(y^2) = arcsin(x^2) + C
Given the initial condition y(0) = 1:
arcsin(1) = arcsin(0) + C
π/2 = C
Therefore, the particular solution to the differential equation is:
arcsin(y^2) = arcsin(x^2) + π/2