y^'=y+x

Lösung

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

y = - x - 1 + c_{1} e^{x}

Schritte zur Lösung

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

Solve linear ODE:

y = - x - 1 + c_{1} e^{x}

y = - x - 1 + c_{1} e^{x}

d er i v a t i v e \frac{72000 + 60 x}{x}

d er i v a t i v e f (x) = cos (sin (x))

t^{2} y^{^{'}} - y = 0

\int \frac{e ^{u}}{( 2 - e ^{u} ) ^{2}} d u

d er i v a t i v e \frac{1}{x} sin^{3} (6 x)