(dy)/(dx)=((x-3))/((y+1))

解答

\frac{d y}{d x} = \frac{( x - 3 )}{( y + 1 )}

y = - 1 + x^{2} - 6 x + c_{1} + 1, y = - x^{2} - 6 x + c_{1} + 1 - 1

求解步骤

\frac{d y}{d x} = \frac{( x - 3 )}{( y + 1 )}

求解可分离方程:

y = - 1 + x^{2} - 6 x + c_{1} + 1, y = - x^{2} - 6 x + c_{1} + 1 - 1

y = - 1 + x^{2} - 6 x + c_{1} + 1, y = - x^{2} - 6 x + c_{1} + 1 - 1

d er i v a t i v e f (x) = \frac{1}{( 1 + x ^{2} )}

d er i v a t i v e \frac{x ^{2} - 3 x + 2}{- x ^{2} - 4 x - 4}

\frac{\partial}{\partial x} (ln (sin (x)))

\int x^{2} (x^{3} + 1)^{3} d x

\int (x^{2} + x) e^{- 2 x} d x