Question

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

Answer 1

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)`