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

Answer 1

Paisley Johnson

#y = 4e^ (x^3/3) - 1#

#dy/dx=x^2(1+y)#

This is separable #1/(1+y) dy/dx=x^2#

#int \ 1/(1+y) dy/dx \ dx = int \ x^2 \ dx#

#int \ 1/(1+y) \ dy = int \ x^2 \ dx#

#ln (1+y) = x^3/3 + C#

#1+y = e^ (x^3/3 + C) = e^ (x^3/3)e^C = Ce^ (x^3/3) #

#y = Ce^ (x^3/3) - 1#

Applying the IV:

#3 = C e^0 -1 = C - 1 implies C = 4#

#y = 4e^ (x^3/3) - 1#

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Answer 2

Charles Arocha

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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Answer from HIX Tutor

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.