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}

\frac{d y}{d x} = \frac{( e ^{x} )}{y 1 + y ^{2}}

\int sin^{3} (3 x) cos^{2} (3 x) d x

t an g e n t f (x) = \frac{6 x}{x + 3}, (- 1, - 3)

\int e^{x} sin (3 x) d x

\frac{\partial}{\partial x} (sin (\frac{2 x}{y}))