y^'=yln(x)+y

Lösung

y^{^{'}} = y ln (x) + y

y = c_{1} x^{x}

Schritte zur Lösung

y^{'} (x) = y ln (x) + y

Solve separable ODE:

y = c_{1} x^{x}

y = c_{1} x^{x}

5 \frac{d x}{d t} + 4 x = 3

- \frac{2 y ^{^{'}}}{y} = k

y^{^{'}} y + x = 1

y^{^{'}} = \frac{y}{81 t ^{2} + 1}

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