Question

Solve the differential equation xdy-(3y+x5(y)1/3)dx=0 .

Answer 1

Answer on Question #40387, Math, Differential Calculus

Solve the differential equation:

$$x dy - (3y + x^3 y^{1/3}) dx = 0$$

Solution:

This equation is called the Bernoulli’s equation.

Rewrite this equation

$$xy' - 3y = x^5 y^{1/3}$$

Divide both sides by $\frac{3}{2}xy^{1/3}$

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

Let $z(x) = y^{\frac{2}{3}}$, which gives $\frac{dz}{dx} = \frac{\frac{2}{3}y'}{y^{\frac{1}{3}}}$. Then we have equation:

$$\frac{dz}{dx} - 2\frac{z}{x} = \frac{2}{3}x^4$$

Let $\mu(x) = e^{\int_{-\frac{2}{x}}^{-\frac{2}{x}} dx} = \frac{1}{x^2}$

Multiply both sides by $\mu(x)$

$$\frac{1}{x^2} \frac{dz}{dx} - 2 \frac{1}{x^2} \frac{z}{x} = \frac{2}{3} \frac{1}{x^2} x^4 = \frac{2}{3} x^2$$

Apply the reverse product rule $g\frac{df}{dx} + f\frac{dg}{dx} = \frac{d}{dx}(fg)$ on the left hand side:

$$\frac{d}{dx} \left(\frac{z}{x^2}\right) = \frac{2x^2}{3}$$

Integrate both sides with respect to $x$:

$$\int \frac{d}{dx} \left(\frac{z}{x^2}\right) dx = \int \frac{2x^2}{3} dx$$

$\frac{z}{x^2} = \frac{2x^3}{9} + C$, where $C$ is an arbitrary constant.

$$z = x^2 \left(\frac{2x^3}{9} + C\right)$$ $$y = z^{\frac{3}{2}} = \left(x^2 \left(\frac{2x^3}{9} + C\right)\right)^{\frac{3}{2}}$$

Answer:

$$y = \left(x^2 \left(\frac{2x^3}{9} + C\right)\right)^{\frac{3}{2}}$$