xy^'-y=x

Lösung

x y^{^{'}} - y = x

y = ln (x) x + c_{1} x

Schritte zur Lösung

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

Solve linear ODE:

y = ln (x) x + c_{1} x

y = ln (x) x + c_{1} x

(x \cdot sin (x))^{^{'}}

\frac{d}{d x} (4 - x^{2} - y^{2})

d er i v a t i v e f (x) = sin (x) + \frac{1}{2} cot (x)

\int_{- 2}^{0} (2 + 4 - x^{2}) d x

\int \frac{4 x - 11}{2 x ^{2} + 7 x - 4} d x