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

Lösung

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

y = e^{- 2 t + 2}

Schritte zur Lösung

\frac{1}{2} y^{'} (t) + y = 0

Solve separable ODE:

y = e^{- 2 t + 2}

y = e^{- 2 t + 2}

t an g e n t f (x) = 1 - x^{2}, (1, 0)

\frac{d}{d x} (\frac{( 6 x ^{2} + 6 x + 8 )}{x})

\frac{\partial}{\partial y} (x 1 + y^{3})

l a pl a ce t r an s f or m 4 t \cdot sin (3 t)

\frac{x y ^{^{'}}}{y ^{2}} sin (\frac{x}{y}) = \frac{1}{y} sin (\frac{x}{y})