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Beliebt Rechnen >

ax^'+bx=0,x(0)=e^{rt}

Lösung

a x^{^{'}} + b x = 0, x (0) = e^{r (x, t) t}

Lösung

a x^{'}

Schritte zur Lösung

Finde

a x^{^{'}} + b x \circ x (0)

gegeben

a x^{^{'}} + b x = 0, x (0) (x, t) = e^{r (x, t) t}

Vereinfache

a x^{^{'}} + b x \circ x \cdot 0 : a x^{^{'}}

= a x^{^{'}}

Graph

Beliebte Beispiele

(4x^5-16x^3-2x)(1-y^2)-2y(x^2-4)y^'=0

(4 x^{5} - 16 x^{3} - 2 x) (1 - y^{2}) - 2 y (x^{2} - 4) y^{^{'}} = 0

derivative of x^2+derivative of y^2=1

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

y^'=(1+x)/(xy),y(1)=-4

y^{^{'}} = \frac{1 + x}{x y}, y (1) = - 4

(dv)/(dt)=0.4v+1000

\frac{d v}{d t} = 0.4 v + 1000

(dy)/y =-(dx)/x

\frac{d y}{y} = - \frac{d x}{x}