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

Lösung

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

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

Schritte zur Lösung

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

Solve linear ODE:

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

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

(e^{5 x})^{^{'}}

x \to 2 + lim ((x - 3)^{2})

\int \frac{x ^{2} - 5 x + 6}{( 2 x - 1 ) ( x - 2 ) ^{2}} d x

x \to \infty lim (x^{3} - 3)

\frac{d}{d x} (10 cos (2 x))