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

Lösung

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

y = - \frac{W ( \frac{2 e ^{x} ( s i n ( x ) - c o s ( x ) ) + c _{1}}{e} )}{2} - \frac{1}{2}

Schritte zur Lösung

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

Solve separable ODE:

y = - \frac{W ( \frac{2 e ^{x} ( s i n ( x ) - c o s ( x ) ) + c _{1}}{e} )}{2} - \frac{1}{2}

y = - \frac{W ( \frac{2 e ^{x} ( s i n ( x ) - c o s ( x ) ) + c _{1}}{e} )}{2} - \frac{1}{2}

y^{^{'}} + 0.5 y = y^{3}

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

\frac{d y}{d x} = - 4 x^{5} e^{y}

(x^{2} + 1) \frac{d y}{d x} = 6 x - 3 x y

y^{^{'}} + x y = \frac{x}{y ^{3}}