y^{''}+4y^'=sin^3(x)

Lösung

y^{^{''}} + 4 y^{^{'}} = sin^{3} (x)

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

Schritte zur Lösung

y^{''} (x) + 4 y^{'} (x) = sin^{3} (x)

Solve linear ODE:

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

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

(d - 1)^{2} (d^{2} + 1) y = 3 x^{2} - 5 x + 2

y^{^{''}} - 4 y = (x^{2} - 4) sin (2 x)

(d^{2} - 13 d + 12) y = 2008

(d^{2} + 9) y = sec (3 x)

(d^{2} - 3 d + 2) \cdot y = e^{5 x}