ty^'+2y+t=0

Lösung

t y^{^{'}} + 2 y + t = 0

y = - \frac{t}{3} + \frac{c _{1}}{t ^{2}}

Schritte zur Lösung

t y^{'} (t) + 2 y + t = 0

Solve linear ODE:

y = - \frac{t}{3} + \frac{c _{1}}{t ^{2}}

y = - \frac{t}{3} + \frac{c _{1}}{t ^{2}}

\int + cos (θ) cos^{5} (sin (θ)) d θ

x^{^{'}} = \frac{1}{x ^{2}}, x (1) = 1

d er i v a t i v e y = \frac{1}{a ^{2} - x ^{2}}

\frac{\partial}{\partial x} (e^{\frac{x}{b ( x )}})

t \to 0 lim (\frac{5}{1 + t})