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Beliebt Rechnen >

2y^'+y=5,y(3)=0

Lösung

2 y^{^{'}} + y = 5, y (3) = 0

Lösung

y = - 5 e^{\frac{- t + 3}{2}} + 5

Schritte zur Lösung

2 y^{'} (t) + y = 5

Solve separable ODE:

y = - 5 e^{\frac{- t + 3}{2}} + 5

y = - 5 e^{\frac{- t + 3}{2}} + 5

Graph

Beliebte Beispiele

xy^2(dy)/(dx)=x+6

x y^{2} \frac{d y}{d x} = x + 6

(dy)/(dx)+5y=-4

\frac{d y}{d x} + 5 y = - 4

y^'=(-(3y^2+1)x)/(x^2+1)

y^{^{'}} = \frac{- ( 3 y ^{2} + 1 ) x}{x ^{2} + 1}

derivative of 7x^2+8ln(xy)=14

\frac{d}{d x} (7 x^{2} + 8 ln (x y)) = 14

y^'=(4x)/(1+2y),y(2)=0

y^{^{'}} = \frac{4 x}{1 + 2 y}, y (2) = 0