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Beliebt Rechnen >

y^{''}+y=3cos(4t),y(0)=1,y^'(0)=2

Lösung

y^{^{''}} + y = 3 cos (4 t), y (0) = 1, y^{^{'}} (0) = 2

Lösung

y = \frac{6}{5} cos (t) + 2 sin (t) - \frac{1}{5} cos (4 t)

Schritte zur Lösung

y^{''} (t) + y = 3 cos (4 t)

Solve linear ODE:

y = \frac{6}{5} cos (t) + 2 sin (t) - \frac{1}{5} cos (4 t)

y = \frac{6}{5} cos (t) + 2 sin (t) - \frac{1}{5} cos (4 t)

Graph

Beliebte Beispiele

x^{''}+x= 1/(cos(t))

x^{^{''}} + x = \frac{1}{cos ( t )}

y^{''}-6y^'+10y=e^{3x}cos(x)

y^{^{''}} - 6 y^{^{'}} + 10 y = e^{3 x} cos (x)

y^{''}+y=x^2+sin(x)

y^{^{''}} + y = x^{2} + sin (x)

y^{''}-4y^'-21y=320e^{-t}

y^{^{''}} - 4 y^{^{'}} - 21 y = 320 e^{- t}

6y^{''}-y^'-y=50e^{x/2}+4x^2-5

6 y^{^{''}} - y^{^{'}} - y = 50 e^{\frac{x}{2}} + 4 x^{2} - 5