(dy)/(dx)=((-y^2))/((1+y^2))*(e^{-x})

Lösung

\frac{d y}{d x} = \frac{( - y ^{2} )}{( 1 + y ^{2} )} \cdot (e^{- x})

y = \frac{c _{1} e ^{x} + 1 + ( - c _{1} e ^{x} - 1 ) ^{2} + 4 e ^{2 x}}{2 e ^{x}}, y = \frac{c _{1} e ^{x} + 1 - ( - c _{1} e ^{x} - 1 ) ^{2} + 4 e ^{2 x}}{2 e ^{x}}

Schritte zur Lösung

\frac{d y}{d x} = \frac{( - y ^{2} )}{( 1 + y ^{2} )} \cdot (e^{- x})

Solve separable ODE:

y = \frac{c _{1} e ^{x} + 1 + ( - c _{1} e ^{x} - 1 ) ^{2} + 4 e ^{2 x}}{2 e ^{x}}, y = \frac{c _{1} e ^{x} + 1 - ( - c _{1} e ^{x} - 1 ) ^{2} + 4 e ^{2 x}}{2 e ^{x}}

y = \frac{c _{1} e ^{x} + 1 + ( - c _{1} e ^{x} - 1 ) ^{2} + 4 e ^{2 x}}{2 e ^{x}}, y = \frac{c _{1} e ^{x} + 1 - ( - c _{1} e ^{x} - 1 ) ^{2} + 4 e ^{2 x}}{2 e ^{x}}

\int \frac{e ^{3 x}}{x} d x

t an g e n t f (x) = \frac{x}{x + 1}

x \to - 2 lim (\frac{1}{( x + 2 ) ^{5}})

in v erse l a pl a ce \frac{1}{s ( s ^{2} + 2 s + 3 )}

\int \frac{2 x + 3}{x ^{2} + 9} d x