5y^'=ycos(5x),y(0)=1

Lösung

5 y^{^{'}} = y cos (5 x), y (0) = 1

y = e^{\frac{s i n ( 5 x )}{25}}

Schritte zur Lösung

5 y^{'} (x) = y cos (5 x)

Solve separable ODE:

y = e^{\frac{s i n ( 5 x )}{25}}

y = e^{\frac{s i n ( 5 x )}{25}}

y^{^{'}} x - y = 0

ln (y^{2}) \frac{d y}{d x} = x^{2} y, y (1) = e^{2}

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

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

y^{^{'}} + 5 y = 0, y (0) = 1