Solve dy:y=x^3-3x^2+2sinx

$$d y : y = x ^ { 3 } - 3 x ^ { 2 } + 2 \sin x$$

Solve for d

$$d=2\sin(x)+x^{3}-3x^{2}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by y.

$$dy=yx^{3}-3x^{2}y+2\sin(x)y$$

The equation is in standard form.

$$yd=2y\sin(x)+yx^{3}-3yx^{2}$$

Divide both sides by y.

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

Dividing by y undoes the multiplication by y.

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

Divide y\left(x^{3}-3x^{2}+2\sin(x)\right) by y.

$$d=2\sin(x)+x^{3}-3x^{2}$$