Question

Решить дифференциальное уравнение методом вариации произвольных постоянных

Answer 1

$y''+4y=8ctg2x\\\\1)\; \; k^2+4=0\; ,\; \; k^2=-4\; ,\; \; k_{1,2}=\pm 2i\\\\y_{obsh.odn.}=C_1^*\, cos2x+C_2^*\, sin2x\\\\2)\; \; y_{obsh.neodn.}=C_1(x)\cdot \underbrace {cos2x}_{y_1}+C_2(x)\cdot \underbrace {sin2x}_{y_2}\\\\\\\left \{ {{C_1'(x)\cdot y_1+C_2'(x)\cdot y_2=0} \atop {C_1'(x)\cdot y_1'+C_2'(x)\cdot y_2'=f(x)}} \right. \; \; \left \{ {{C_1'(x)\cdot cos2x+C_2'(x)\cdot sin2x=0} \atop {C_1'(x)\cdot (-2sin2x)+C_2'(x)\cdot 2cos2x=8ctg2x}} \right.$

$\Delta =\left|\begin{array}{cc}cos2x&sin2x\\-2sin2x&2cos2x\end{array}\right|=2cos^22x+2sin^22x=2\ne 0\\\\\Delta _1=\left|\begin{array}{cc}0&sin2x\\8ctg2x&2cos2x\end{array}\right|=-8ctg2x\cdot sin2x=-8cos2x\\\\\Delta _2=\left|\begin{array}{cc}cos2x&0\\-2sin2x&8ctg2x\end{array}\right|=8ctg2x\cdot cos2x=\frac{8cos^2x}{sin2x}\\\\\\C_1'(x)=\frac{\Delta _1}{\Delta } =-4cos2x\; ,\; \; C_2'(x)=\frac{\Delta _2(x)}{\Delta }=\frac{4cos^22x}{sin2x}\\\\C_1(x)=\int (-4cos2x)dx=-2sin2x+C_1\\\\C_2(x)=\int \frac{4cos^22x}{sin2x}dx=\int \frac{4(1-sin^22x)}{sin2x}dx=4\int (\frac{1}{sin2x}-sin2x)dx=$

$=4\int (\frac{1}{2sinx\, cosx}-sin2x)dx=4\int (\frac{1}{2\cdot \frac{sinx}{cosx}\cdot cos^2x}-sin2x)dx=\\\\=2\int \frac{d(tgx)}{tgx}-\frac{4}{2}\int sin2x\, d(2x)=2ln|tgx|-2(-cos2x)+C_2$

$y=C_1(x)\cdot cos2x+C_2(x)\cdot sin2x=\\\\=(-2sin2x+C_1)\cdot cos2x+(2ln|tgx|+2cos2x+C_2)\cdot sin2x=\\\\=C_1\, cos2x+C_2\, sin2x-2sin2x\cdot cos2x+2sin2x\cdot ln|tgx|+\\\\+2cos2x\cdot sin2x\; ,\\\\y=C_1\, cos2x+C_2\, sin2x+2\, sin2x\cdot ln|tgx|$