t^2x^'+x=0

Lösung

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

x = e^{\frac{1}{t} + c_{1}}

Schritte zur Lösung

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

Solve separable ODE:

x = e^{\frac{1}{t} + c_{1}}

x = e^{\frac{1}{t} + c_{1}}

\frac{d y}{d x} = \frac{e ^{5 x}}{cos ( 2 y )}

y^{^{'}} - 2 x^{2} y - y = 0

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

y y^{^{'}} = [sin (x) - cos (x)] 1 - y^{2}

\frac{d y}{d x} = 13 x sec (y)