(2x+3y-1)dx-4(x+1)dy=0

Lösung

(2 x + 3 y - 1) d x - 4 (x + 1) d y = 0

y = \frac{2 ( 3 x + 4 )}{3} + \frac{1}{3} + c_{1} (x + 1)^{\frac{3}{4}}

Schritte zur Lösung

(2 x + 3 y - 1) d x - 4 (x + 1) d y = 0

Solve linear ODE:

y = \frac{2 ( 3 x + 4 )}{3} + \frac{1}{3} + c_{1} (x + 1)^{\frac{3}{4}}

y = \frac{2 ( 3 x + 4 )}{3} + \frac{1}{3} + c_{1} (x + 1)^{\frac{3}{4}}

in v erse l a pl a ce \frac{e ^{- 2 s}}{s ( s - 2 )}

\frac{d}{d x} (- 2 e^{5 y} x - 2 sin (2 x))

\int_{- 2}^{2} (4 - x^{2}) d x

t a y l or (1 - x^{2})^{\frac{1}{2}}

d er i v a t i v e 6 x^{\frac{1}{2}}