e^y*y^'=t+t^3

Lösung

e^{y} \cdot y^{^{'}} = t + t^{3}

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

Schritte zur Lösung

e^{y} \cdot y^{'} (t) = t + t^{3}

Solve separable ODE:

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

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

\frac{( x ^{2} - 9 ) d y}{d x} + x y = 0

\frac{d y}{d x} = 3 y (1 + y)

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

\frac{d v}{d t} = 1 - 7 v

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