y^{''}+a^2y^'=cos(bx)

Lösung

y^{^{''}} + a^{2} y^{^{'}} = cos (b x)

y = c_{1} + c_{2} e^{- a^{2} x} + \frac{a ^{2} sin ( b x )}{b ( b ^{2} + a ^{4} )} - \frac{cos ( b x )}{b ^{2} + a ^{4}}

Schritte zur Lösung

y^{''} (x) + a^{2} y^{'} (x) = cos (b x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- a^{2} x} + \frac{a ^{2} sin ( b x )}{b ( b ^{2} + a ^{4} )} - \frac{cos ( b x )}{b ^{2} + a ^{4}}

y = c_{1} + c_{2} e^{- a^{2} x} + \frac{a ^{2} sin ( b x )}{b ( b ^{2} + a ^{4} )} - \frac{cos ( b x )}{b ^{2} + a ^{4}}

y^{^{''}} - 5 y^{^{'}} + 4 y = 2 x^{2}

y^{^{''}} + 16 y = 7 x, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} + 4 y = 3 x (2 + cos (2 x))

l in e a r y^{^{''}} - 2 y^{^{'}} + 5 y = 6 e^{t}, y (0) = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = 10 u, y (0) = 0, y^{^{'}} (0) = 0