Solve dy/dx+p(x)y=Q(x)/y^2

$$\frac { d y } { d x } + p ( x ) y = \frac { Q ( x ) } { y ^ { 2 } }$$

Solve for Q

$$\left\{\begin{matrix}Q=py^{3}\text{, }&x\neq 0\text{ and }y\neq 0\\Q\in \mathrm{R}\text{, }&x=0\text{ and }y\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by y^{2}.

$$y^{2}\frac{\mathrm{d}}{\mathrm{d}x}(y)+pxyy^{2}=Qx$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

Swap sides so that all variable terms are on the left hand side.

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

The equation is in standard form.

$$xQ=pxy^{3}$$

Divide both sides by x.

$$\frac{xQ}{x}=\frac{pxy^{3}}{x}$$

Dividing by x undoes the multiplication by x.

$$Q=\frac{pxy^{3}}{x}$$

Divide pxy^{3} by x.

$$Q=py^{3}$$

Solve for p

$$\left\{\begin{matrix}p=\frac{Q}{y^{3}}\text{, }&x\neq 0\text{ and }y\neq 0\\p\in \mathrm{R}\text{, }&x=0\text{ and }y\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by y^{2}.

$$y^{2}\frac{\mathrm{d}}{\mathrm{d}x}(y)+pxyy^{2}=Qx$$

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

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

Subtract y^{2}\frac{\mathrm{d}}{\mathrm{d}x}(y) from both sides.

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

Reorder the terms.

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

The equation is in standard form.

$$xy^{3}p=Qx$$

Divide both sides by xy^{3}.

$$\frac{xy^{3}p}{xy^{3}}=\frac{Qx}{xy^{3}}$$

Dividing by xy^{3} undoes the multiplication by xy^{3}.

$$p=\frac{Qx}{xy^{3}}$$

Divide Qx by xy^{3}.

$$p=\frac{Q}{y^{3}}$$