How do you use the differential equation #dy/dx=4x+(9x^2)/(3x^3+1)^(3/2)# to find the equation of the function given point (0,2)?

1 Answer
Noah G
Jan 10, 2017

#y = 2x^2 - 2/sqrt(3x^3 + 1) + 4#

Explanation:

This is a separable differential equation.

#dy/dx = 4x + (9x^2)/(3x^3 + 1)^(3/2)#

#dy = (4x + (9x^2)/(3x^3 + 1)^(3/2))dx#

Integrate both sides.

#int(dy) = int 4x + (9x^2)/(3x^3 + 1)^(3/2)dx#

We can integrate the right-hand side using the rule #intx^ndx= x^(n + 1)/(n + 1) + C#, where #n != -1# and a substitution. Let #u = 3x^3 + 1#. Then #du = 9x^2dx# and #dx = (du)/(9x^2)#.

#y = 2x^2 + int (9x^2)/u^(3/2) * (du)/(9x^2)#

#y = 2x^2 + intu^(-3/2)du#

Integrate using the rule above:

#y = 2x^2 + -2/u^(1/2)#

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

The last step is to find the value of #C#. We know that when #x = 0#, #y = 2#. So:

#2 = 2(0)^2 - 2/sqrt(3(0)^3 + 1)+ C#

#2 = -2 + C#

#C = 4#

The solution to the differential equation is therefore #y = 2x^2 - 2/sqrt(3x^3 + 1) + 4#. Differentiating will yield the original equation.

Hopefully this helps!

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