(4-t^2)y^'+2ty=3t^2

Lösung

(4 - t^{2}) y^{^{'}} + 2 t y = 3 t^{2}

y = - \frac{3}{2} ln (t + 2) + \frac{3}{8} t^{2} ln (t + 2) - \frac{3}{t + 2} + \frac{3 t ^{2}}{4 ( t + 2 )} + \frac{3}{2} ln (t - 2) - \frac{3}{8} t^{2} ln (t - 2) - \frac{3}{t - 2} + \frac{3 t ^{2}}{4 ( t - 2 )} + 4 c_{1} - c_{1} t^{2}

Schritte zur Lösung

(4 - t^{2}) y^{'} (t) + 2 t y = 3 t^{2}

Solve linear ODE:

y = - \frac{3}{2} ln (t + 2) + \frac{3}{8} t^{2} ln (t + 2) - \frac{3}{t + 2} + \frac{3 t ^{2}}{4 ( t + 2 )} + \frac{3}{2} ln (t - 2) - \frac{3}{8} t^{2} ln (t - 2) - \frac{3}{t - 2} + \frac{3 t ^{2}}{4 ( t - 2 )} + 4 c_{1} - c_{1} t^{2}

y = - \frac{3}{2} ln (t + 2) + \frac{3}{8} t^{2} ln (t + 2) - \frac{3}{t + 2} + \frac{3 t ^{2}}{4 ( t + 2 )} + \frac{3}{2} ln (t - 2) - \frac{3}{8} t^{2} ln (t - 2) - \frac{3}{t - 2} + \frac{3 t ^{2}}{4 ( t - 2 )} + 4 c_{1} - c_{1} t^{2}

a re a f (x) = 2 sin (x) + x, - π, π

\int_{1}^{e} x^{3} \cdot ln (x) d x

\frac{\partial}{\partial z} (x y + z^{3} x - 2 yz)

\frac{d}{d x} (ln (\frac{3 x + 4}{3 x - 4}))

\int sin (t) 2 t d t