Solve dx^2/dx+8xy=7-x^4)

$$\frac { d x ^ { 2 } } { d x } + 8 x y = 7 - x ^ { 4 } )$$

Solve for y

$$y=-\frac{x^{3}}{8}-\frac{1}{4}+\frac{7}{8x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$8xy=7-x^{4}-\frac{\mathrm{d}}{\mathrm{d}x}(x^{2})$$

Reorder the terms.

$$8xy=-\frac{\mathrm{d}}{\mathrm{d}x}(x^{2})-x^{4}+7$$

The equation is in standard form.

$$8xy=7-2x-x^{4}$$

Divide both sides by 8x.

$$\frac{8xy}{8x}=\frac{7-2x-x^{4}}{8x}$$

Dividing by 8x undoes the multiplication by 8x.

$$y=\frac{7-2x-x^{4}}{8x}$$

Divide -2x-x^{4}+7 by 8x.

$$y=-\frac{x^{3}}{8}-\frac{1}{4}+\frac{7}{8x}$$