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Beliebt Rechnen >

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

Lösung

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

Lösung

y = e^{\frac{t}{2}} (- cos (\frac{t}{2}) + 5 sin (\frac{t}{2}))

Schritte zur Lösung

2 y^{''} (t) - 2 y^{'} (t) + y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (- cos (\frac{t}{2}) + 5 sin (\frac{t}{2}))

y = e^{\frac{t}{2}} (- cos (\frac{t}{2}) + 5 sin (\frac{t}{2}))

Graph

Beliebte Beispiele

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

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

(d^2y)/(dt^2)+16y=0

\frac{d ^{2} y}{d t ^{2}} + 16 y = 0

(d^2x)/(dt^2)+2(dx)/(dt)+8x=0

\frac{d ^{2} x}{d t ^{2}} + 2 \frac{d x}{d t} + 8 x = 0

(d^2x)/(dt^2)+4(dx)/(dt)+8x=0

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 8 x = 0

x*y^{''}+2x*y^'+x*y=0

x \cdot y^{^{''}} + 2 x \cdot y^{^{'}} + x \cdot y = 0