# What is a solution to the differential equation #dy/dx=y^2+1# with y(1)=0?

grouping variables

Integrating both sides

then

Now, observing the initial condition

and concluding

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The solution to the differential equation ( \frac{dy}{dx} = y^2 + 1 ) with the initial condition ( y(1) = 0 ) is ( y(x) = \tan(x) ).

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To solve the given initial value problem, we can use the method of separation of variables.

Given the differential equation ( \frac{dy}{dx} = y^2 + 1 ), we separate variables:

[ \frac{dy}{y^2 + 1} = dx ]

Now, integrate both sides:

[ \int \frac{dy}{y^2 + 1} = \int dx ]

The integral on the left side can be solved using the arctangent function:

[ \arctan(y) = x + C ]

Where ( C ) is the constant of integration.

Now, to find the value of the constant ( C ), we use the initial condition ( y(1) = 0 ):

[ \arctan(0) = 1 + C ] [ 0 = C + \frac{\pi}{4} ] [ C = -\frac{\pi}{4} ]

Therefore, the particular solution to the given differential equation with the initial condition ( y(1) = 0 ) is:

[ \arctan(y) = x - \frac{\pi}{4} ]

Or, equivalently:

[ y = \tan(x - \frac{\pi}{4}) ]

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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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