Question

Решите дифференциальное уравнение.

Answer 1

$y''-y'+9y=sinx\\\\a)\ \ k^2-k+9=0\ \ ,\ \ \ D=1-36=-35\ \ ,\ \ x_{1,2}=\dfrac{1\pm i\sqrt{35}}{2}=\dfrac{1}{2}\pm \dfrac{\sqrt{35}}{2}\, i\\\\\\y_{obshee}=e^{0,5x}\cdot \Big(C_1\cdot cos\dfrac{\sqrt{35}}{2}\, x+C_2\cdot sin\dfrac{\sqrt{35}}{2}\, x\Big)$

$b)\ \ f(x)=sinx\ \ ,\\\\y_{chastnoe}=A\, cosx+B\, sinx\ \ \ \Big|\cdot 1\\\\y_{chastnoe}'=-A\, sinx+B\, cosx\ \Big|\cdot (-1)\\\\y_{chastnoe}''=-A\, cosx-B\, sinx\ \Big|\cdot 9\\\\y''-y'+9y=A\, cosx+B\, sinx+A\, sinx-B\, cosx-9A\, cosx-9B\, sinx\ =\ sinx\ ,\\\\(-8A-B)\, cosx+(-8B+A)\, sinx=sinx\ ,\\\\cosx\ |\ -8A-B=0\ ,\ \ B=-8A\ ,\\sinx\ |-8B+A=1\ ,\ \ \ 64A+A=1\ ,\ \ 65A=1\ ,\ A=\dfrac{1}{65}\ ,\ B=-\dfrac{8}{65}$

$y_{chastnoe}=\dfrac{1}{65}\, cosx-\dfrac{8}{65}\, sinx\\\\y=e^{0,5x}\cdot \Big(C_1\cdot cos\dfrac{\sqrt{35}}{2}\, x+C_2\cdot sin\dfrac{\sqrt{35}}{2}\, x\Big)+\dfrac{1}{65}\, cosx-\dfrac{8}{65}\, sinx$