x^'=(1+t)/(t^2x^2)

Lösung

x^{^{'}} = \frac{1 + t}{t ^{2} x ^{2}}

x = 3 \frac{- 3 + 3 t ln ( t ) + 3 c _{1} t}{t}

Schritte zur Lösung

x^{'} (t) = \frac{1 + t}{t ^{2} x ^{2}}

Solve separable ODE:

x = 3 \frac{- 3 + 3 t ln ( t ) + 3 c _{1} t}{t}

x = 3 \frac{- 3 + 3 t ln ( t ) + 3 c _{1} t}{t}

x (y - 1) d x + (x + 1) d y = 0

3 y^{^{'}} + 3 x y = 3 x y^{- 2}

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

y^{^{'}} = - 2 y e^{t}, y (0) = 1

4 \frac{d y}{d x} = 1 + 2 x + 3 y^{2} + 6 x y^{2}