y^{''}=2y+tan^3(x),y(0)= 1/2 tan(x)

Lösung

y^{^{''}} = 2 y + tan^{3} (x), y (0) = \frac{1}{2} tan (x)

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int \frac{2 e ^{- 2 x} tan ^{3} ( x )}{4} d x + e^{- 2 x} \cdot \int - \frac{2 e ^{2 x} tan ^{3} ( x )}{4} d x

Schritte zur Lösung

y^{''} (x) = 2 y + tan^{3} (x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int \frac{2 e ^{- 2 x} tan ^{3} ( x )}{4} d x + e^{- 2 x} \cdot \int - \frac{2 e ^{2 x} tan ^{3} ( x )}{4} d x

y = c_{1} e^{2 x} + c_{2} e^{- 2 x} + e^{2 x} \cdot \int \frac{2 e ^{- 2 x} tan ^{3} ( x )}{4} d x + e^{- 2 x} \cdot \int - \frac{2 e ^{2 x} tan ^{3} ( x )}{4} d x

y^{^{''}} - 7 y^{^{'}} + 10 y = 30 t + 9, y (0) = 7, y^{^{'}} (0) = 26

- 4 y^{^{''}} + y^{^{'}} + 9 y = 15

y^{^{''}} + 50 y^{^{'}} + 100 = 0

y^{^{''}} - 3 y^{^{'}} = - 18 x, y (0) = 0, y^{^{'}} (0) = 5

y^{^{''}} - 8 y^{^{'}} + 17 y = x^{2} e^{4 x}