y^'=y-2x

Lösung

y^{^{'}} = y - 2 x

y = 2 (x + 1) + c_{1} e^{x}

Schritte zur Lösung

y^{'} (x) = y - 2 x

Solve linear ODE:

y = 2 (x + 1) + c_{1} e^{x}

y = 2 (x + 1) + c_{1} e^{x}

\int \frac{4 x ^{3}}{1 + x ^{4}} d x

\frac{\partial}{\partial x} (x^{2} + 2)

\int_{1}^{8} \frac{5 ln ( x )}{x ^{2}} d x

n = 0 \sum \infty n - 2

y^{^{''}} + 2 y^{^{'}} + 3 y = 0, y^{^{'}} (0) = 2, y (0) = 1