y^'=y+e^x,y(0)=(x+c)e^x,y(0)= 1/2

Lösung

y^{^{'}} = y + e^{x}, y (0) = (x + c) e^{x}, y (0) = \frac{1}{2}

y = e^{x} x + c_{1} e^{x}

Schritte zur Lösung

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

Solve linear ODE:

y = e^{x} x + c_{1} e^{x}

y = e^{x} x + c_{1} e^{x}

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

y^{^{'}} = 2 y + (x^{2}) + 5

(x - 1) y^{^{'}} + x y = sin (x)

y^{^{'}} - y = 2 \cdot x \cdot e^{2 x}

y^{^{'}} + y + e^{- x} = 0