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Beliebt Rechnen >

y^{''}+25y=40cos(5t),y(0)=y

Lösung

y^{^{''}} + 25 y = 40 cos (5 t), y (0) = y

Lösung

y = c_{1} cos (5 t) + c_{2} sin (5 t) + 4 t sin (5 t)

Schritte zur Lösung

y^{''} (t) + 25 y = 40 cos (5 t)

Solve linear ODE:

y = c_{1} cos (5 t) + c_{2} sin (5 t) + 4 t sin (5 t)

y = c_{1} cos (5 t) + c_{2} sin (5 t) + 4 t sin (5 t)

Graph

Beliebte Beispiele

y^{''}-4y^'+8y=e^{2x}+sin(2x)

y^{^{''}} - 4 y^{^{'}} + 8 y = e^{2 x} + sin (2 x)

y^{''}+4y^'-21y=5e^{2x}+4

y^{^{''}} + 4 y^{^{'}} - 21 y = 5 e^{2 x} + 4

y^{''}+4y=1-t^3

y^{^{''}} + 4 y = 1 - t^{3}

(d^2y)/(dx^2)-(dy)/(dx)-2y=sin(2x)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = sin (2 x)

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

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