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

Lösung

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

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

Schritte zur Lösung

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

Solve separable ODE:

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

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

y^{^{'}} = (x + 1) e^{- x} y^{2}

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

t y^{^{'}} = 2 y

x y \frac{d y}{d x} = 1 - x^{2}

(y + 1) y^{^{'}} - x^{2} y - y = 0, y (3) = 1