(\frac{dy)/(dt)}{sqrt(y)}=((2t+1))/(t^2)

Lösung

\frac{\frac{d y}{d t}}{y} = \frac{( 2 t + 1 )}{t ^{2}}

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

Schritte zur Lösung

\frac{\frac{d y}{d t}}{y} = \frac{( 2 t + 1 )}{t ^{2}}

Solve separable ODE:

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

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

\frac{x y ^{^{'}}}{y} = 2

\frac{d x}{d t} = x^{2} + \frac{1}{9}

\frac{d y}{d x} = x y^{2} + y^{2} + x + 1

- 3 y^{^{'}} - 6 y = 0

x d x = - \frac{ey}{ey + 6} d y