# What is a particular solution to the differential equation #dy/dx=2xsqrt(1-y^2)# and y(0)=0?

this is separable

meaning

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To find a particular solution to the differential equation ( \frac{dy}{dx} = 2x\sqrt{1-y^2} ) with the initial condition ( y(0) = 0 ), we can separate variables and then integrate.

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

Integrate both sides: [ \int \frac{dy}{\sqrt{1-y^2}} = \int 2x , dx ]

The left integral can be solved using a trigonometric substitution ( y = \sin(\theta) ), which gives: [ \int \frac{dy}{\sqrt{1-y^2}} = \int \frac{d\theta}{\cos(\theta)} = \int \sec(\theta) , d\theta = \ln|\sec(\theta) + \tan(\theta)| + C ]

where ( C ) is the constant of integration.

Now, for the right integral: [ \int 2x , dx = x^2 + D ]

where ( D ) is another constant of integration.

Combining the results and simplifying, we get: [ \ln|\sec(\theta) + \tan(\theta)| + C = x^2 + D ]

Applying the initial condition ( y(0) = 0 ) gives ( \theta = 0 ), which simplifies the left side to ( \ln(1) = 0 ).

Therefore, the particular solution to the differential equation is ( y = \sin(\theta) = x^2 ).

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