Solve dy/dx=x/v+2y^3

$$\frac { d y } { d x } = \frac { x } { v } + 2 y ^ { 3 }$$

Solve for v

$$\left\{\begin{matrix}v=-\frac{x}{2y^{3}}\text{, }&x\neq 0\text{ and }y\neq 0\\v\neq 0\text{, }&x=0\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Variable v cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by v.

$$v\frac{\mathrm{d}}{\mathrm{d}x}(y)=x+2y^{3}v$$

Subtract 2y^{3}v from both sides.

$$v\frac{\mathrm{d}}{\mathrm{d}x}(y)-2y^{3}v=x$$

Combine all terms containing v.

$$\left(\frac{\mathrm{d}}{\mathrm{d}x}(y)-2y^{3}\right)v=x$$

The equation is in standard form.

$$\left(-2y^{3}\right)v=x$$

Divide both sides by -2y^{3}.

$$\frac{\left(-2y^{3}\right)v}{-2y^{3}}=\frac{x}{-2y^{3}}$$

Dividing by -2y^{3} undoes the multiplication by -2y^{3}.

$$v=\frac{x}{-2y^{3}}$$

Divide x by -2y^{3}.

$$v=-\frac{x}{2y^{3}}$$

Variable v cannot be equal to 0.

$$v=-\frac{x}{2y^{3}}\text{, }v\neq 0$$

Solve for x

$$x=-2vy^{3}$$

$$v\neq 0$$