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Beliebt Rechnen >

ax^'+bx=0

Lösung

a x^{^{'}} + b x = 0

Lösung

x = e^{\frac{- b t + c _{1}}{a}}

Schritte zur Lösung

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

Solve separable ODE:

x = e^{\frac{- b t + c _{1}}{a}}

x = e^{\frac{- b t + c _{1}}{a}}

Graph

Beliebte Beispiele

(x^2+x-6)y+(x+3)y^'+(x-2)y=0

(x^{2} + x - 6) y + (x + 3) y^{^{'}} + (x - 2) y = 0

y - 3 y^{^{'}} - 18 = 0

(dy)/(dt)y=(ln(6t))^2

\frac{d y}{d t} y = (ln (6 t))^{2}

(dy)/(dt)=-2(t+1)y^2,y(0)=-1/8

\frac{d y}{d t} = - 2 (t + 1) y^{2}, y (0) = - \frac{1}{8}

(ye^x+xye^x)dx=-(xe^x+2)dy

(y e^{x} + x y e^{x}) d x = - (x e^{x} + 2) d y