Question

Найти общее решение дифференциального уравнения.

Answer 1

$y''+25y=-10sin5x+20cos5x+50e^{5x}\\\\1)\ \ y''+25y=0\ \ \ \to \ \ \ k^2+25=0\ \ ,\ \ k^2=-25\ ,\ \ k_{1,2}=\pm 5i\\\\y_{obshee\ odnor.}=C_1cos5x+C_2sin5x\\\\2)\ \ f(x)=f_1(x)+f_2(x)\ \ ,\ \ f_1(x)=20cos5x-10sin5x\ ,\ \ f_2(x)=50e^{5x}\ \ ,\\\\y_1=x\cdot (Acos5x+Bsin5x)\\\\y_1'=Acos5x+Bsin5x+x\cdot (-5Asin5x+5Bcos5x)\\\\y_1''=-5Asin5x+5Bcos5x-5Asin5x+5Bcos5x+x\cdot (-25Acos5x-25Bsin5x)$

$y_1''+25y_1=-5Asin5x+5Bcos5x-5Asin5x+5Bcos5x+\\\\+x\cdot (-25Acos5x-25Bsin5x)+25x\cdot (Acos5x+Bsin5x)=-10sin5x+20cos5x$

$cos5x\ |\ 10B=20\ \ ,\qquad \qquad B=2\ ,\\sin5x\ |\ -10A=-10\ \ ,\qquad A=1\ .\\\\y_1=x\cdot (cos5x+2sin5x)\\\\\\y_2=D\, e^{5x}\\\\y_2'=5De^{5x}\\\\y_2''=25De^{5x}\\\\y''_2+25y_2=25De^{5x}+25De^{5x}=50De^{5x}\ \ ,\ \ 50De^{5x}=50e^{5x}\ \ ,\ \ D=1\\\\y_2=e^{5x}$

$3)\ \ y_{obshee\ neodn.}=C_1cos5x+C_2sin5x+x\, (cos5x+2sin5x)+e^{5x}$