Solve ∫ (x^2-4y) wrt ydx=1

$$\int ( x ^ { 2 } - 4 y ) d y d x = 1$$

Solve for d

$$d=\frac{1}{x\left(С+yx^{2}-2y^{2}\right)}$$

$$\left(y\neq 0\text{ and }x\neq 0\text{ and }|x|\neq \frac{\sqrt{8y+\frac{С}{y}}}{2}\right)\text{ or }\left(x\neq 0\text{ and }С_{1}<0\text{ and }y<0\right)\text{ or }\left(x\neq 0\text{ and }С_{2}\geq 0\text{ and }y<\sqrt{С_{3}}\text{ and }y>0\right)\text{ or }\left(С_{4}\neq 0\text{ and }x\neq 0\text{ and }|x|\neq \frac{\sqrt{8y+\frac{С_{5}}{y}}}{2}\right)\text{ or }\left(С_{6}\neq 0\text{ and }x\neq 0\text{ and }y=0\right)\text{ or }\left(x\neq 0\text{ and }y\leq 0\text{ and }y<\sqrt{С_{7}}\text{ and }С_{8}<0\right)\text{ or }\left(x\neq 0\text{ and }y<\sqrt{С_{9}}\text{ and }y\geq 0\text{ and }С_{10}>0\right)\text{ or }\left(x\neq 0\text{ and }y<-\sqrt{С_{11}}\right)$$

Steps for Solving Linear Equation

The equation is in standard form.

$$x\left(С+yx^{2}-2y^{2}\right)d=1$$

Divide both sides by \left(x^{2}y-2y^{2}+С\right)x.

$$\frac{x\left(С+yx^{2}-2y^{2}\right)d}{x\left(С+yx^{2}-2y^{2}\right)}=\frac{1}{x\left(С+yx^{2}-2y^{2}\right)}$$

Dividing by \left(x^{2}y-2y^{2}+С\right)x undoes the multiplication by \left(x^{2}y-2y^{2}+С\right)x.

$$d=\frac{1}{x\left(С+yx^{2}-2y^{2}\right)}$$