# What is a solution to the differential equation #dy/dx=x^2(1+y)# with y=3 when x=0?

Applying the IV:

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To find the solution, we integrate both sides of the equation:

∫(1+y) dy = ∫x^2 dx

Integrating, we get:

y + 0.5*y^2 = (1/3) * x^3 + C

Given that y = 3 when x = 0, we substitute these values:

3 + 0.5*3^2 = (1/3) * 0^3 + C 3 + 4.5 = 0 + C C = 7.5

So the particular solution to the differential equation is:

y + 0.5*y^2 = (1/3) * x^3 + 7.5

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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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