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

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 $\frac{\mathrm{dy}}{\mathrm{dx}}$ for y':

$y \sqrt{1 - {x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} - x \left(1 + {y}^{2}\right) = 0$

Move the second term to the right side:

$y \sqrt{1 - {x}^{2}} \frac{\mathrm{dy}}{\mathrm{dx}} = x \left(1 + {y}^{2}\right)$

Multiply both sides of the equation by $\frac{\mathrm{dx}}{\sqrt{1 - {x}^{2}} \left(1 + {y}^{2}\right)}$

$\frac{y}{1 + {y}^{2}} \mathrm{dy} = \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

Integrate both sides:

$\int \frac{y}{1 + {y}^{2}} \mathrm{dy} = \int \frac{x}{\sqrt{1 - {x}^{2}}} \mathrm{dx}$

$\left(\frac{1}{2}\right) \ln \left(1 + {y}^{2}\right) = - \sqrt{1 - {x}^{2}} + C$

Multiply both sides by 2:

$\ln \left(1 + {y}^{2}\right) = - 2 \sqrt{1 - {x}^{2}} + C$

Use the exponential function:

$1 + {y}^{2} = {e}^{- 2 \sqrt{1 - {x}^{2}} + C}$

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

$1 + {y}^{2} = C {e}^{- 2 \sqrt{1 - {x}^{2}}}$

Subtract 1 from both sides:

${y}^{2} = C {e}^{- 2 \sqrt{1 - {x}^{2}}} - 1$

Square root both sides:

$y = \pm \sqrt{C {e}^{- 2 \sqrt{1 - {x}^{2}}} - 1} \text{ [1]}$

Use the point to solve for C, substitute 0 for x and $\sqrt{3}$ for y:

$\sqrt{3} = \pm \sqrt{C {e}^{- 2 \sqrt{1 - {0}^{2}}} - 1}$

The above equation can only be true, if we can drop the $\pm$ and make it only positive:

$\sqrt{3} = \sqrt{C {e}^{- 2 \sqrt{1 - {0}^{2}}} - 1}$

Square both sides and solve for C:

$3 = C {e}^{-} 2 - 1$

$4 = C {e}^{-} 2$

$C = 4 {e}^{2}$

Substitute $4 {e}^{2}$ for C and remove the $\pm$ in equation [1]:

$y = \sqrt{\left(4 {e}^{2}\right) {e}^{- 2 \sqrt{1 - {x}^{2}}} - 1}$