4y^{''}+4y+y=3xe^x

Lösung

4 y^{^{''}} + 4 y + y = 3 x e^{x}

y = c_{1} cos (\frac{5 x}{2}) + c_{2} sin (\frac{5 x}{2}) + \frac{e ^{x} x}{3} - \frac{8}{27} e^{x}

Schritte zur Lösung

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

Solve linear ODE:

y = c_{1} cos (\frac{5 x}{2}) + c_{2} sin (\frac{5 x}{2}) + \frac{e ^{x} x}{3} - \frac{8}{27} e^{x}

y = c_{1} cos (\frac{5 x}{2}) + c_{2} sin (\frac{5 x}{2}) + \frac{e ^{x} x}{3} - \frac{8}{27} e^{x}

y^{^{''}} + 4 y + 4 = 0

x^{^{''}} + 8 x^{^{'}} + 16 x = e^{- t} sin (4 t)

y^{^{''}} + y^{^{'}} - y = sin (x)

y^{^{''}} + y^{^{'}} = 4 t

y^{^{'''}} + y^{^{''}} = 4 x^{2} + 3 e^{x}