y^{''}-10y^'+25=e^{5x}ln(4+x^2)

Lösung

y^{^{''}} - 10 y^{^{'}} + 25 = e^{5 x} ln (4 + x^{2})

y = c_{1} e^{10 x} + c_{2} + e^{10 x} \cdot \int \frac{e ^{5 x} ln ( 4 + x ^{2} ) - 25}{10 e ^{10 x}} d x + \int - \frac{e ^{5 x} ln ( x ^{2} + 4 ) - 25}{10} d x

Schritte zur Lösung

y^{''} (x) - 10 y^{'} (x) + 25 = e^{5 x} ln (4 + x^{2})

Solve linear ODE:

y = c_{1} e^{10 x} + c_{2} + e^{10 x} \cdot \int \frac{e ^{5 x} ln ( 4 + x ^{2} ) - 25}{10 e ^{10 x}} d x + \int - \frac{e ^{5 x} ln ( x ^{2} + 4 ) - 25}{10} d x

y = c_{1} e^{10 x} + c_{2} + e^{10 x} \cdot \int \frac{e ^{5 x} ln ( 4 + x ^{2} ) - 25}{10 e ^{10 x}} d x + \int - \frac{e ^{5 x} ln ( x ^{2} + 4 ) - 25}{10} d x

y^{^{''}} + y = 4 x^{2}

y^{^{'''}} + 4 y^{^{'}} = sec (x) cos (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 6 e^{t}, y (0) = 1, y^{^{'}} (0) = 2

y^{^{'''}} + y^{^{''}} + y^{^{'}} + y = e^{- t} + 15 t

y^{^{''}} + 9 y = e^{x}