3x^2y^2dx+(2y^3x+4y^3)dy=0

解答

3 x^{2} y^{2} d x + (2 y^{3} x + 4 y^{3}) d y = 0

y = \frac{- 3 x ^{2} + 12 x - 24 ln ( x + 2 ) + 36 + c _{1}}{2}, y = - \frac{- 3 x ^{2} + 12 x - 24 ln ( x + 2 ) + 36 + c _{1}}{2}

求解步骤

3 x^{2} y^{2} d x + (2 y^{3} x + 4 y^{3}) d y = 0

求解可分离方程:

y = \frac{- 3 x ^{2} + 12 x - 24 ln ( x + 2 ) + 36 + c _{1}}{2}, y = - \frac{- 3 x ^{2} + 12 x - 24 ln ( x + 2 ) + 36 + c _{1}}{2}

y = \frac{- 3 x ^{2} + 12 x - 24 ln ( x + 2 ) + 36 + c _{1}}{2}, y = - \frac{- 3 x ^{2} + 12 x - 24 ln ( x + 2 ) + 36 + c _{1}}{2}

5 x^{^{''}} + 25 x^{^{'}} + 1700 x = 0

\frac{1}{y ^{\frac{1}{3}}} d y = 1 d x

\int - 4 sin (2 x) + 3 cos (3 x) d x

x \to 0 + lim (x sin (\frac{1}{x}))

d er i v a t i v e P (t) = 0.45 t^{2} - 12 t + 105