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Beliebt Rechnen >

y^{''}+y^'=cos(2t),y(0)=0,y^'(0)=1

Lösung

y^{^{''}} + y^{^{'}} = cos (2 t), y (0) = 0, y^{^{'}} (0) = 1

Lösung

y = \frac{4}{5} + \frac{1}{5} - \frac{4}{5} e^{- t} + \frac{1}{10} sin (2 t) - \frac{1}{5} cos (2 t)

Schritte zur Lösung

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

Solve linear ODE:

y = \frac{4}{5} + \frac{1}{5} - \frac{4}{5} e^{- t} + \frac{1}{10} sin (2 t) - \frac{1}{5} cos (2 t)

y = \frac{4}{5} + \frac{1}{5} - \frac{4}{5} e^{- t} + \frac{1}{10} sin (2 t) - \frac{1}{5} cos (2 t)

Graph

Beliebte Beispiele

y^{''}-2y^'+1=e^x

y^{^{''}} - 2 y^{^{'}} + 1 = e^{x}

y^{''}-4y=cos^2(x)

y^{^{''}} - 4 y = cos^{2} (x)

y^{''}+9y=5csc^2(3x)

y^{^{''}} + 9 y = 5 csc^{2} (3 x)

y^{''}-y-12y=e^{4x}

y^{^{''}} - y - 12 y = e^{4 x}

(d^2y)/(dx^2)-y= 1/x

\frac{d ^{2} y}{d x ^{2}} - y = \frac{1}{x}