Question

МАТЕМАТИКА

ТЕРМІНОВО

ДОПОМОЖІТЬ ,БУДЬ ЛАСКА

розв'язати задачу Коші методом варіації сталої

(завдання на фото )​

Answer 1

Ответ:

Пошаговое объяснение:

$y''+\pi ^2=\frac{\pi ^2}{sin(\pi x)}\\1)\ \\y''+\pi ^2y=0\\ k^2+\pi ^2=0\\k^2=-\pi ^2\\k=б\pi i\ \ \ \ \alpha =0\ \ \ \ \ \beta =\pi \ \ \ \ \ \Rightarrow\\y=e^{\alpha x}(C_1sin(\beta x)+C_2cos(\beta x))\\y=e^{0*x}C_1sin(\pi x)+C_2cos(\pi x)\\y(\frac{1}{2} )=C_1sin\frac{\pi }{2}+C_2cos\frac{\pi }{2}=1\\ C_1*1+C_2*0=1\\ C_1=1.$

$y'=(C_1sin(\pi x)+C_2cos(\pi x))'=C_1\pi cos(\pi x)-C_2\pi sin(\pi x)\\y'(\frac{1}{2} )=C_1\pi cos\frac{\pi }{2}-C_2sin\frac{\pi }{2}=\frac{\pi ^2}{2} \\C_1*0-C_2*1=\frac{\pi ^2}{2}\\ C_2=-\frac{\pi ^2}{2} .\ \ \ \ \Rightarrow\\y_0=sin(\pi x)-\frac{\pi ^2}{2} cos(\pi x).$

$2)\\y^*=\frac{\pi ^2}{sin(\pi x)} \\y^{'*}=(\frac{\pi ^2}{sin(\pi x)})'=\pi ^2*(\frac{1}{sin(\pi x)} )'=\pi ^2*(-\frac{\pi*cos(\pi x)}{sin^2(\pi x)})=-\frac{\pi ^3cos(\pi x)}{sin^2(\pi x)}.\\ y^{''*}=(-\frac{\pi ^3cos(\pi x)}{sin^2(\pi x)} )'=-\pi ^3*(\frac{cos(\pi x)}{sin^2(\pi x)})'=-\pi ^3*\frac{-\pi *sin^3(\pi x)+2\pi *cos^2(\pi x)*sin\pi x}{sin^4(\pi x)}=\\ =\pi ^4*\frac{sin^2(\pi x)+2cos^2(\pi x)}{sin^3(\pi x)}.\ \ \ \ \Rightarrow$

$\pi ^4*\frac{sin^2(\pi x)+2cos^2(\pi x)}{sin^3(\pi x)} +\pi ^2*\frac{\pi ^2}{sin(\pi x)}=\frac{A*\pi ^2}{sin(\pi x)}\ |:\pi ^2$

$\frac{\pi ^2*(sin^2(\pi x)+2cos^2(\pi x)}{sin^3(\pi x)} +\frac{\pi ^2}{sin(\pi x)}=\frac{A}{sin(\pi x)}\\A=\pi ^2.\ \ \ \ \ \Rightarrow\\y=sin(\pi x)-\frac{\pi^2 }{2}*cos(\pi x)+\frac{\pi ^2}{sin(\pi x)} .$