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

Grouping variables

so it is a separable differential equation.

We have also

so

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To find the particular solution to the differential equation y√(1-x^2)y' - x√(1-y^2) = 0 with the initial condition y(0) = 1, we first rewrite the equation in terms of separation of variables. Then, we integrate both sides and solve for y to find the particular solution. After solving the equation, we substitute the initial condition y(0) = 1 to find the specific value of the constant of integration.

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

- Separate variables by moving terms involving (y) to one side and terms involving (x) to the other side.
- Integrate both sides with respect to their respective variables.
- Solve for (y).
- Apply the initial condition (y(0) = 1) to find the particular solution.

Let's solve the equation:

[y\sqrt{1-x^2}y' = x\sqrt{1-y^2}]

[y\sqrt{1-y^2} , dy = x\sqrt{1-x^2} , dx]

[\int y\sqrt{1-y^2} , dy = \int x\sqrt{1-x^2} , dx]

After integrating both sides, we'll have:

[-\frac{1}{2}(1-y^2)^{3/2} = -\frac{1}{2}(1-x^2)^{3/2} + C]

Now, solve for (y):

[1-y^2 = (1-x^2) + C]

[y^2 = 1 - (1-x^2) - C]

[y^2 = x^2 + C]

Apply the initial condition (y(0) = 1):

[1^2 = 0 + C]

[C = 1]

So, the particular solution is:

[y^2 = x^2 + 1]

[y = \sqrt{x^2 + 1}]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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