y^{''}+4y=sin(t)\H(t-2pi)

Lösung

y^{^{''}} + 4 y = sin (t) H (t - 2 π)

y = c_{1} cos (2 t) + c_{2} sin (2 t) + cos (2 t) \cdot \int - \frac{sin ( 2 t ) sin ( t ) H ( t - 2 π )}{2} d t + sin (2 t) \cdot \int \frac{cos ( 2 t ) sin ( t ) H ( t - 2 π )}{2} d t

Schritte zur Lösung

y^{''} (t) + 4 y = sin (t) H (t - 2 π)

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + cos (2 t) \cdot \int - \frac{sin ( 2 t ) sin ( t ) H ( t - 2 π )}{2} d t + sin (2 t) \cdot \int \frac{cos ( 2 t ) sin ( t ) H ( t - 2 π )}{2} d t

y = c_{1} cos (2 t) + c_{2} sin (2 t) + cos (2 t) \cdot \int - \frac{sin ( 2 t ) sin ( t ) H ( t - 2 π )}{2} d t + sin (2 t) \cdot \int \frac{cos ( 2 t ) sin ( t ) H ( t - 2 π )}{2} d t

4 y^{^{''}} + 16 y = 1

14 y^{^{''}} + y^{^{'}} + y = x^{2} - 3 x

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = x^{3} e^{2 x}

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

y^{^{'''}} + 16 y^{^{'}} = e^{- π x}