y^'=3x+y

Lösung

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

y = - 3 (x + 1) + c_{1} e^{x}

Schritte zur Lösung

y^{'} (x) = 3 x + y

Solve linear ODE:

y = - 3 (x + 1) + c_{1} e^{x}

y = - 3 (x + 1) + c_{1} e^{x}

\frac{d ^{3}}{d x ^{3}} (ln (x + 1))

s l o p e (0, 2), (4, - 2)

81 y^{^{''}} + 18 y^{^{'}} + y = 0

s l o p e (1.2) (2.4)

\frac{d}{d x} (\frac{4 x ^{3} + 3 x ^{2}}{x})