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

Lösung

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

y = \frac{2 ( - arctan ( tan ( \frac{x}{2} ) ) ( tan ( \frac{x}{2} ) - 1 ) - 1 )}{( tan ( \frac{x}{2} ) - 1 ) ( tan ( x ) + sec ( x ) )} + \frac{c _{1}}{tan ( x ) + sec ( x )}

Schritte zur Lösung

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

Solve linear ODE:

y = \frac{2 ( - arctan ( tan ( \frac{x}{2} ) ) ( tan ( \frac{x}{2} ) - 1 ) - 1 )}{( tan ( \frac{x}{2} ) - 1 ) ( tan ( x ) + sec ( x ) )} + \frac{c _{1}}{tan ( x ) + sec ( x )}

y = \frac{2 ( - arctan ( tan ( \frac{x}{2} ) ) ( tan ( \frac{x}{2} ) - 1 ) - 1 )}{( tan ( \frac{x}{2} ) - 1 ) ( tan ( x ) + sec ( x ) )} + \frac{c _{1}}{tan ( x ) + sec ( x )}

l a pl a ce t r an s f or m e^{3 t} (t^{2} + 4)

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\int_{1}^{2} sin (\frac{nπ x}{2}) d x