Question

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)?

Answer 1

`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!