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

Lösung

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

y = c_{1} cos (t) + c_{2} sin (t) - \frac{t ^{3} cos ( t )}{6} + \frac{t ^{2} sin ( t )}{4} - \frac{t ^{2} cos ( t )}{4} + \frac{t sin ( t )}{4} + \frac{t cos ( t )}{4}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} cos (t) + c_{2} sin (t) - \frac{t ^{3} cos ( t )}{6} + \frac{t ^{2} sin ( t )}{4} - \frac{t ^{2} cos ( t )}{4} + \frac{t sin ( t )}{4} + \frac{t cos ( t )}{4}

y = c_{1} cos (t) + c_{2} sin (t) - \frac{t ^{3} cos ( t )}{6} + \frac{t ^{2} sin ( t )}{4} - \frac{t ^{2} cos ( t )}{4} + \frac{t sin ( t )}{4} + \frac{t cos ( t )}{4}

t an g e n t f (x) = 0.5 x^{2} + 2 x - 3, at x = - 3

\int \frac{1}{z ^{3} z ^{2} - 4} d z

y^{^{'}} + 3 y = 2 e^{- 3 t}, y (0) = 2

\int 3 x (x - 4) d x

\frac{d}{d x} (x^{2} z (x))