xy^'=2y+3x

Lösung

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

y = - 3 x + c_{1} x^{2}

Schritte zur Lösung

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

Solve linear ODE:

y = - 3 x + c_{1} x^{2}

y = - 3 x + c_{1} x^{2}

t an g e n t y = 5 x - x^{2} (1.4)

t an g e n t y = 8 x^{4}, at x = 0

t an g e n t f (x) = 3 + ln (x)

\frac{d}{d x} (- sin (3 x) \cdot 3)

\int \frac{x ^{3}}{x ^{2} - 4} d x