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

Question

Luke Alvarez · Accepted Answer

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

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)#

Scarlett Allison · Answer

undefined To find the particular solution to the differential equation $ y\sqrt{1-x^2}y' - x(1+y^2) = 0 $ that satisfies $ y(0) = \sqrt{3} $, follow these steps:

1. Rewrite the differential equation in standard form.
2. Separate variables and integrate both sides.
3. Solve for $ y $ to find the particular solution.
4. Use the initial condition $ y(0) = \sqrt{3} $ to determine the constant of integration.

Step 1: Rewrite the differential equation:
$$ y\sqrt{1-x^2} \frac{dy}{dx} = x(1+y^2) $$

Step 2: Separate variables and integrate:
$$ \int \frac{y}{1+y^2} \, dy = \int \frac{x}{\sqrt{1-x^2}} \, dx $$

Step 3: Solve for $ y $:
$$ \frac{1}{2} \ln(1+y^2) = -\frac{1}{2} \arcsin(x^2) + C $$

$$ \ln(1+y^2) = -\arcsin(x^2) + C_1 $$

$$ 1 + y^2 = e^{-\arcsin(x^2) + C_1} $$

$$ y^2 = e^{-\arcsin(x^2) + C_1} - 1 $$

$$ y = \pm \sqrt{e^{-\arcsin(x^2) + C_1} - 1} $$

Step 4: Apply the initial condition $ y(0) = \sqrt{3} $ to find $ C_1 $:
$$ \sqrt{3} = \sqrt{e^{-\arcsin(0)} - 1} $$
$$ \sqrt{3} = \sqrt{e^0 - 1} $$
$$ \sqrt{3} = \sqrt{1} $$
$$ C_1 = 0 $$

Therefore, the particular solution satisfying $ y(0) = \sqrt{3} $ is:
$$ y = \sqrt{e^{-\arcsin(x^2)}} $$