y^'=(2t)/(y+t^2y)

Lösung

y^{^{'}} = \frac{2 t}{y + t ^{2} y}

y = 2 ln (1 + t^{2}) + c_{1}, y = - 2 ln (1 + t^{2}) + c_{1}

Schritte zur Lösung

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

Solve separable ODE:

y = 2 ln (1 + t^{2}) + c_{1}, y = - 2 ln (1 + t^{2}) + c_{1}

y = 2 ln (1 + t^{2}) + c_{1}, y = - 2 ln (1 + t^{2}) + c_{1}

n \to \infty lim (\frac{1}{n !})

x \to 0 lim (\frac{1}{x ^{2} ( x + 8 )})

y^{^{''}} + 4 y = sin (2 x)

\int_{0}^{8} 3 x d x

\frac{d y}{d x} + 4 y^{2} = 0