y^{''}+10y^'+25y=e^{(-10x)/(x^2)}

Lösung

y^{^{''}} + 10 y^{^{'}} + 25 y = e^{\frac{- 10 x}{x ^{2}}}

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + e^{- 5 x} \cdot \int - e^{\frac{5 x ^{2} - 10}{x}} x d x + e^{- 5 x} x \cdot \int e^{\frac{5 x ^{2} - 10}{x}} d x

Schritte zur Lösung

y^{''} (x) + 10 y^{'} (x) + 25 y = e^{\frac{- 10 x}{x ^{2}}}

Solve linear ODE:

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + e^{- 5 x} \cdot \int - e^{\frac{5 x ^{2} - 10}{x}} x d x + e^{- 5 x} x \cdot \int e^{\frac{5 x ^{2} - 10}{x}} d x

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + e^{- 5 x} \cdot \int - e^{\frac{5 x ^{2} - 10}{x}} x d x + e^{- 5 x} x \cdot \int e^{\frac{5 x ^{2} - 10}{x}} d x

d (r) = \frac{( ln ( 2 ) )}{ln ( 1 + 0.04 )}

y^{^{''}} + y^{^{'}} = 2 x - 2

2 y^{^{''}} - 4 y^{^{'}} - 6 y = 16 e^{- x}

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} + y = x^{- 2} e^{x}

x^{^{''}} + 2 x^{^{'}} + 5 x = sin (2 t)