y^'+1/(x+8)y=x^{-2},y(1)=5

Lösung

y^{^{'}} + \frac{1}{x + 8} y = x^{- 2}, y (1) = 5

y = \frac{ln ( x )}{x + 8} - \frac{8}{x ( x + 8 )} + \frac{53}{x + 8}

Schritte zur Lösung

y^{'} (x) + \frac{1}{x + 8} y = x^{- 2}

Solve linear ODE:

y = \frac{ln ( x )}{x + 8} - \frac{8}{x ( x + 8 )} + \frac{53}{x + 8}

y = \frac{ln ( x )}{x + 8} - \frac{8}{x ( x + 8 )} + \frac{53}{x + 8}

\frac{d}{d x} (\frac{( x + 1 ) ^{3}}{x ^{\frac{3}{2}}})

t an g e n t f (x) = 4 x - x^{2}, at x = 1

x \to - 2 - lim (\frac{x ^{2} - 4}{x + 2})

\int x x^{2} - 2 d x

x \to 0 lim (\frac{π}{x})