yy^'=sin(x)+cos(x)

Lösung

y y^{^{'}} = sin (x) + cos (x)

y = - 2 cos (x) + 2 sin (x) + c_{1}, y = - - 2 cos (x) + 2 sin (x) + c_{1}

Schritte zur Lösung

y y^{'} (x) = sin (x) + cos (x)

Solve separable ODE:

y = - 2 cos (x) + 2 sin (x) + c_{1}, y = - - 2 cos (x) + 2 sin (x) + c_{1}

y = - 2 cos (x) + 2 sin (x) + c_{1}, y = - - 2 cos (x) + 2 sin (x) + c_{1}

y^{^{'}} = y sin (x)

y y^{^{'}} = x - x^{2}, y (0) = x 1 - x^{2}

\frac{d y}{d x} - \frac{y ^{2}}{x} = 0

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

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