4x+5yy^'=3x^2+4y^3y^'

Lösung

4 x + 5 y y^{^{'}} = 3 x^{2} + 4 y^{3} y^{^{'}}

y = - \frac{- 5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{4}, y = - - \frac{- 5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{4}, y = \frac{5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{2}, y = - \frac{5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{2}

Schritte zur Lösung

4 x + 5 y y^{'} (x) = 3 x^{2} + 4 y^{3} y^{'} (x)

Solve separable ODE:

y = - \frac{- 5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{4}, y = - - \frac{- 5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{4}, y = \frac{5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{2}, y = - \frac{5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{2}

y = - \frac{- 5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{4}, y = - - \frac{- 5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{4}, y = \frac{5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{2}, y = - \frac{5 + 25 + 8 ( - c _{1} + 4 x ^{2} - 2 x ^{3} )}{2}

x d y + 2 y d x = 0

y^{^{'}} + e^{x^{2}} y = 0

\frac{d y}{d x} = y x^{2} - 1.1 y

\frac{d y}{d t} = 2 - 3 y

2 y^{^{'}} + y = 0, y (0) = e^{- \frac{x}{2}}