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)?

Jul 24, 2017

The function has equation

$y = {\left(2 {x}^{3} + 1\right)}^{3} - 23$

Explanation:

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 $18 {x}^{2}$? This condition will allow the substitution to be effective.

Let $u = 2 {x}^{3} + 1$. Then $\mathrm{du} = 6 {x}^{2} \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{6 {x}^{2}}$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 18 {x}^{2} \left({u}^{2}\right) \cdot \frac{\mathrm{du}}{6 {x}^{2}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 3 {u}^{2} \mathrm{du}$

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

$\int \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \int \left(3 {u}^{2}\right) \mathrm{du}$

Now use $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$

$y = {u}^{3} + C$

You can now reverse the substitution.

$y = {\left(2 {x}^{3} + 1\right)}^{3} + C$

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

$4 = {\left(2 {\left(1\right)}^{3} + 1\right)}^{3} + C$

$4 = {\left(2 + 1\right)}^{3} + C$

$4 - 27 = C$

$C = - 23$

Hopefully this helps!