4y=y^'

Lösung

4 y = y^{^{'}}

y = e^{4 t + c_{1}}

Schritte zur Lösung

4 y = y^{'} (t)

Solve separable ODE:

y = e^{4 t + c_{1}}

y = e^{4 t + c_{1}}

x y d x + d y = 0

(sin (x)) d y = 2 y (cos (x)) d x

- (x y^{^{'}} + y) (csc (2) (x y) - 1) = 0

y^{^{'}} = 0.2 (y - 100)

x^{4} y^{^{'}} + y^{3} x^{3} = 0