Zbadać czy problem początkowy posiada rozwiązanie

[maritka210]

W poniższych zadaniach zbadać czy problem początkowy posiada rozwiązanie i czy rozwiązanie jest dokładnie jedno. Jeżeli istnieje rozwiązanie to należy go wyznaczyć. W przypadku gdy jest więcej niż jedno rozwiązanie należy podać przykłady conajmniej dwóch różnych rozwiązań

Nie wiem jak sie za to zabrac nawet.

\(\displaystyle{ y'' + 4y = -5sin(2t) + 3cos(2t), y(0) -1, y'(0) = 1}\)

[bartek118]

To jest równanie liniowe. Najpierw rozwiązujemy równanie
\(\displaystyle{ y'' + 4y = 0}\)
korzystając z wielomianu charakterystycznego \(\displaystyle{ \lambda^2 + 4 = 0}\), a potem korzystamy z metody przewidywań.

[maritka210]

\(\displaystyle{ y'' + 4y = -5sin(2t) + 3cos(2t), y(0) = -1, y'(0) = 1}\)

\(\displaystyle{ \frac{d^{2}y(t)}{dt^{2}} + 4y(t) = 3cos(2t) -5sin(2t)}\)

\(\displaystyle{ \frac{d^{2}y(t)}{dt^{2}} + 4y(t) = 0}\)

\(\displaystyle{ y(t) = e^{\lambda t}}\)

\(\displaystyle{ \frac{d^{2}}{dt^{2}}(e^{\lambda t}) + 4e^{\lambda t} = 0}\)

\(\displaystyle{ \frac{d^{2}}{dt^{2}}(e^{\lambda t}) = \lambda^{2}e^{\lambda t}}\)

\(\displaystyle{ \lambda^{2}e^{\lambda t} + 4e^{\lambda t} = 0}\)

\(\displaystyle{ (\lambda^2 + 4)e^{\lambda t} = 0}\)

\(\displaystyle{ e^{\lambda t} \neq 0}\)

\(\displaystyle{ \lambda^{2} +4 =0}\)

\(\displaystyle{ \lambda = 2i \lor \lambda = -2i}\)

\(\displaystyle{ \lambda = \pm 2i}\)

\(\displaystyle{ y_{1}(t) = c_{1}e^{2it}}\)

\(\displaystyle{ y_{2}(t) = c_{2}e^{-2it}}\)

\(\displaystyle{ y(t) = y_{1}(t) + y_{2}(t) = c_{1}e^{2it} + c_{2}e^{-2it}}\)

\(\displaystyle{ e^{ \alpha +i \beta } = e^{ \alpha }cos( \beta ) +ie^{ \alpha }sin( \beta )}\)

\(\displaystyle{ y(t) = c_{1}(cos(2t)+isin(2t)) + c_{2}(cos(2t)-isin(2t))}\)

\(\displaystyle{ y(t) = ( c_{1} + c_{2} )cos(2t) + i( c_{1} - c_{2} )sin(2t)}\)

\(\displaystyle{ c_{1} = c_{1} + c_{2}}\)

\(\displaystyle{ c_{2} = i( c_{1} - c_{2} )}\)

\(\displaystyle{ y(t) = c_{1}cos(2t) + c_{2}sin(2t)}\)

\(\displaystyle{ \frac{d^{2}y(t)}{dt^{2}} +4y(t) = 3cos(2t) - 5sin(2t)}\)

\(\displaystyle{ y_{p}(t) = t(a_{1}cos(2t) a_{2}sin(2t))}\)

\(\displaystyle{ \frac{d^{2}y_{p}(t)}{dt^{2}} = \frac{d^2}{dt^2}(a_{1}tcos(2t) + a_{2}tsin(2t)) = -4a_{1}tcos(2t) -
4a_{1}sin(2t) + 4a_{2}cos(2t) - 4a_{2}sin(2t)}\)

\(\displaystyle{ \frac{d^{2}y_{p}(t)}{dt^{2}} + 4y_{p}(t) = 3cos(2t) - 5sin(2t) -4a_{1}tcos(2t)-4a_{1}sin(2t)+
4a_{2}cos(2t)-4a_{2}tsin(2t) + 4(a_{1}tcos(2t)+a_{2}tsin(2t)) = 3cos(2t) - 5sin(2t)}\)

\(\displaystyle{ 4a_{2}cos(2t) - 4a_{1}sin(2t) = 3cos(2t) - 5sin(2t)}\)

\(\displaystyle{ 4a_{2} = 3}\)

\(\displaystyle{ -4a_{1} = -5}\)

\(\displaystyle{ a_{1} = \frac{5}{4}}\)

\(\displaystyle{ a_{2} = \frac{3}{4}}\)

\(\displaystyle{ y_{p}(t) = tcos(2t)a_{1} + tsin(2t)a_{2}}\)

\(\displaystyle{ y_{p}(t) = \frac{5}{4}tcos(2t) + \frac{3}{4} tsin(2t)}\)

\(\displaystyle{ y(t) = y_{c}(t) + y_{p}(t) = \frac{5}{4}tcos(2t)a_{1} + \frac{3}{4} tsin(2t) + c_{1}cos(2t) + c_{2}sin(2t)}\)

\(\displaystyle{ \frac{dy(t)}{dt} = \frac{d}{dt} ( \frac{5}{4}tcos(2t) + \frac{3}{4} tsin(2t) + c_{1}cos(2t) + c_{2}sin(2t) )}\)-- 27 gru 2018, o 17:10 --\(\displaystyle{ = \frac{5}{4} cos(2t) + \frac{3}{2} tcos(2t) + \frac{3}{4}sin(2t) - \frac{5}{2} tsin(2t) - 2c_{1}sin(2t) +2c_{2}cos(2t)}\)

\(\displaystyle{ y(0) = -1}\)

\(\displaystyle{ \frac{dy(t)}{dt} = \frac{5}{4} cos(2t) + \frac{3}{2}tcos(2t) + \frac{3}{4} sin(2t) - \frac{5}{2}tsin(2t) - 2sin(2t)c_1 + 2cos(2t)c_{2}}\)

\(\displaystyle{ 2c_{2} + \frac {5}{4} = 1}\)

\(\displaystyle{ c_{1} = -1}\)

\(\displaystyle{ c_{2} = -\frac {1}{8}}\)

\(\displaystyle{ y(t) = \frac{5}{4} tcos(2t) + \frac {3}{4} tsin(2t) + cos(2t)c_{1} + sin(2t)c_{2}}\)

\(\displaystyle{ y(t) = \frac{1}{8}(2cos(2t)(5t - 4) + (6t - 1) sin(2t))}\)

Co o tym myślisz?