y^{''}-8y^'+16y=(e^{-4x})/(1+x^2)

Lösung

y^{^{''}} - 8 y^{^{'}} + 16 y = \frac{e ^{- 4 x}}{1 + x ^{2}}

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + e^{4 x} \cdot \int - \frac{x}{e ^{8 x} ( 1 + x ^{2} )} d x + e^{4 x} x \cdot \int \frac{1}{e ^{8 x} ( 1 + x ^{2} )} d x

Schritte zur Lösung

y^{''} (x) - 8 y^{'} (x) + 16 y = \frac{e ^{- 4 x}}{1 + x ^{2}}

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + e^{4 x} \cdot \int - \frac{x}{e ^{8 x} ( 1 + x ^{2} )} d x + e^{4 x} x \cdot \int \frac{1}{e ^{8 x} ( 1 + x ^{2} )} d x

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + e^{4 x} \cdot \int - \frac{x}{e ^{8 x} ( 1 + x ^{2} )} d x + e^{4 x} x \cdot \int \frac{1}{e ^{8 x} ( 1 + x ^{2} )} d x

y^{^{''''}} - 4 y^{^{''}} = t^{2} - e^{t}

y^{^{''}} - 2 y^{^{'}} + y = \frac{9 e ^{t}}{1 + t ^{2}}

\frac{d ^{2} y}{d x ^{2}} - π^{2} y = x cos (x)

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = - 6 e x + 10 cos (x)

\frac{d ^{3} y}{d x ^{3}} - \frac{d ^{2} y}{d x ^{2}} = e^{x}