y^'=y+x,y(0)=1

Lösung

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

y = - x - 1 + 2 e^{x}

Schritte zur Lösung

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

Solve linear ODE:

y = - x - 1 + 2 e^{x}

y = - x - 1 + 2 e^{x}

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (2) = 2

\int sin (t) e^{- \frac{2}{3} t} d t

\int_{1}^{2} x^{2} e^{x^{3} - 1} d x

n = 1 \sum \infty \frac{1}{n 2}

y^{^{''}} + 9 y = (sec (3 t))^{2}