How do you find the particular solution to ysqrt(1-x^2)y'-x(1+y^2)=0 that satisfies y(0)=sqrt3?

1 Answer
Dec 31, 2016

Use the separation of variables method, integrate both sides, and then use the specified point to evaluate the constant.

Explanation:

Given: ysqrt(1 - x^2)y' - x(1 + y^2) = 0; y(0) = sqrt(3)

Use the notation dy/dx for y':

ysqrt(1 - x^2)dy/dx - x(1 + y^2) = 0

Move the second term to the right side:

ysqrt(1 - x^2)dy/dx = x(1 + y^2)

Multiply both sides of the equation by dx/(sqrt(1 - x^2)(1 + y^2))

y/(1 + y^2)dy = x/sqrt(1 - x^2)dx

Integrate both sides:

inty/(1 + y^2)dy = intx/sqrt(1 - x^2)dx

(1/2)ln(1 + y^2) = -sqrt(1 - x^2) + C

Multiply both sides by 2:

ln(1 + y^2) = -2sqrt(1 - x^2) + C

Use the exponential function:

1 + y^2 = e^(-2sqrt(1 - x^2) + C)

Adding a constant in the exponent is the same as multiplying by a constant:

1 + y^2 = Ce^(-2sqrt(1 - x^2))

Subtract 1 from both sides:

y^2 = Ce^(-2sqrt(1 - x^2)) - 1

Square root both sides:

y = +-sqrt(Ce^(-2sqrt(1 - x^2)) - 1)" [1]"

Use the point to solve for C, substitute 0 for x and sqrt(3) for y:

sqrt(3) = +-sqrt(Ce^(-2sqrt(1 - 0^2)) - 1)

The above equation can only be true, if we can drop the +- and make it only positive:

sqrt(3) = sqrt(Ce^(-2sqrt(1 - 0^2)) - 1)

Square both sides and solve for C:

3 = Ce^-2 - 1

4 = Ce^-2

C = 4e^2

Substitute 4e^2 for C and remove the +- in equation [1]:

y = sqrt((4e^2)e^(-2sqrt(1 - x^2)) - 1)