4y^{''}+4y^'+3y=4e^{-3}x+18x+15

Lösung

4 y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{- 3} x + 18 x + 15

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2})) + \frac{( 4 + 18 e ^{3} ) x}{3 e ^{3}} + \frac{- 27 e ^{3} - 16}{9 e ^{3}}

Schritte zur Lösung

4 y^{''} (x) + 4 y^{'} (x) + 3 y = 4 e^{- 3} x + 18 x + 15

Solve linear ODE:

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2})) + \frac{( 4 + 18 e ^{3} ) x}{3 e ^{3}} + \frac{- 27 e ^{3} - 16}{9 e ^{3}}

y = e^{- \frac{x}{2}} (c_{1} cos (\frac{x}{2}) + c_{2} sin (\frac{x}{2})) + \frac{( 4 + 18 e ^{3} ) x}{3 e ^{3}} + \frac{- 27 e ^{3} - 16}{9 e ^{3}}

(d^{5} - d) y = 12 e^{x} + 8 sin (x) - 2 x

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = e^{x} - x + 16

(d^{3} - 3 d + 2) \cdot y = 540 x^{3} \cdot e^{- x}

(d^{3} + 8) \cdot y = x^{4} + 2 x + 1 + 3 sin (x)

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