Solve dy=2x+xy^2/4y+yx^2

$$d y = \frac { 2 x + x y ^ { 2 } } { 4 y + y x ^ { 2 } }$$

Solve for d

$$d=\frac{x\left(y^{2}+2\right)}{y^{2}\left(x^{2}+4\right)}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by y\left(x^{2}+4\right).

$$dyy\left(x^{2}+4\right)=2x+xy^{2}$$

Multiply y and y to get y^{2}.

$$dy^{2}\left(x^{2}+4\right)=2x+xy^{2}$$

Use the distributive property to multiply dy^{2} by x^{2}+4.

$$dy^{2}x^{2}+4dy^{2}=2x+xy^{2}$$

Combine all terms containing d.

$$\left(y^{2}x^{2}+4y^{2}\right)d=2x+xy^{2}$$

The equation is in standard form.

$$\left(x^{2}y^{2}+4y^{2}\right)d=xy^{2}+2x$$

Divide both sides by y^{2}x^{2}+4y^{2}.

$$\frac{\left(x^{2}y^{2}+4y^{2}\right)d}{x^{2}y^{2}+4y^{2}}=\frac{x\left(y^{2}+2\right)}{x^{2}y^{2}+4y^{2}}$$

Dividing by y^{2}x^{2}+4y^{2} undoes the multiplication by y^{2}x^{2}+4y^{2}.

$$d=\frac{x\left(y^{2}+2\right)}{x^{2}y^{2}+4y^{2}}$$

Divide x\left(2+y^{2}\right) by y^{2}x^{2}+4y^{2}.

$$d=\frac{x\left(y^{2}+2\right)}{y^{2}\left(x^{2}+4\right)}$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{4+4y^{2}+y^{4}-16d^{2}y^{4}}+y^{2}+2}{2dy^{2}}\text{; }x=\frac{-\sqrt{4+4y^{2}+y^{4}-16d^{2}y^{4}}+y^{2}+2}{2dy^{2}}\text{, }&d\neq 0\text{ and }y\neq 0\text{ and }|d|\leq \frac{1}{4}+\frac{1}{2y^{2}}\\x=0\text{, }&d=0\text{ and }y\neq 0\end{matrix}\right.$$