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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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.
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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