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)?

Jan 10, 2017

$y = 2 {x}^{2} - \frac{2}{\sqrt{3 {x}^{3} + 1}} + 4$

Explanation:

This is a separable differential equation.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 x + \frac{9 {x}^{2}}{3 {x}^{3} + 1} ^ \left(\frac{3}{2}\right)$

$\mathrm{dy} = \left(4 x + \frac{9 {x}^{2}}{3 {x}^{3} + 1} ^ \left(\frac{3}{2}\right)\right) \mathrm{dx}$

Integrate both sides.

$\int \left(\mathrm{dy}\right) = \int 4 x + \frac{9 {x}^{2}}{3 {x}^{3} + 1} ^ \left(\frac{3}{2}\right) \mathrm{dx}$

We can integrate the right-hand side using the rule $\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C$, where $n \ne - 1$ and a substitution. Let $u = 3 {x}^{3} + 1$. Then $\mathrm{du} = 9 {x}^{2} \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{9 {x}^{2}}$.

$y = 2 {x}^{2} + \int \frac{9 {x}^{2}}{u} ^ \left(\frac{3}{2}\right) \cdot \frac{\mathrm{du}}{9 {x}^{2}}$

$y = 2 {x}^{2} + \int {u}^{- \frac{3}{2}} \mathrm{du}$

Integrate using the rule above:

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

$y = 2 {x}^{2} - \frac{2}{\sqrt{3 {x}^{3} + 1}} + C$

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

$2 = 2 {\left(0\right)}^{2} - \frac{2}{\sqrt{3 {\left(0\right)}^{3} + 1}} + C$

$2 = - 2 + C$

$C = 4$

The solution to the differential equation is therefore $y = 2 {x}^{2} - \frac{2}{\sqrt{3 {x}^{3} + 1}} + 4$. Differentiating will yield the original equation.

Hopefully this helps!