y^{''}+4y=t^2

Lösung

y^{^{''}} + 4 y = t^{2}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8}

Schritte zur Lösung

y^{''} (t) + 4 y = t^{2}

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8}

d er i v a t i v e f (x) = e^{\infty}

d er i v a t i v e (5 x^{2} - 2 x + 4) (4 x^{2} + 3 x - 6)

\int - x^{2} + 6 d x

x \to 5 lim (\frac{1}{( x - 5 )})

\frac{\partial}{\partial x} (x^{2} \frac{1}{y + 1})