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How do you use the differential equation #dy/dx=18x^2(2x^3+1)^2# to find the equation of the function given point (0,4)?

Answer 1

The function has equation

#y = (2x^3 + 1)^3 - 23#

In this problem, I figure that a substitution would be much simpler than integration by parts. Notice how the binomial within the parentheses has a highest term that is a degree higher than the #18x^2#? This condition will allow the substitution to be effective.

Let #u = 2x^3 + 1#. Then #du = 6x^2dx# and #dx = (du)/(6x^2)#.

#dy/dx = 18x^2(u^2) * (du)/(6x^2)#

#dy/dx = 3u^2 du#

To solve for the function #y#, we must integrate on both side of the equation.

#int (dy/dx) = int (3u^2)du#

Now use #int x^ndx = x^(n + 1)/(n + 1) + C#

#y = u^3 + C#

You can now reverse the substitution.

#y = (2x^3 + 1)^3 + C#

All that is left to do is to solve for #C#.

#4 = (2(1)^3 + 1)^3 + C#

#4 = (2 + 1)^3 + C#

#4 - 27 = C#

#C = -23#

Hopefully this helps!

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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.