Question

Найти решение задачи Коши, используя метод операционного
исчисления.

Answer 1

Объяснение:

$\left \{ {{x'-4y=8sin2t,\ \ \ \ x(0)=-2} \atop {y'-x=-8sin2t,\ \ \ \ y(0)=2}} \right. \\y(t) \ \ \rightarrow\ \ Y(p)\\y'(t)\ \ \rightarrow\ \ Y(p)*p-y(0)=Y(p)*p-2.\\x(t)\ \ \rightarrow\ \ X(p)\\x'(t)\ \ \rightarrow\ \ X(p)*p-x(0)=X(p)-(-2)=X(p)+2.\\\left \{ {{X(p)*p-4*Y(p)=8*\frac{2}{p^2+2^2} } \atop {Y(p)*p-2-X(p)=-8*\frac{2}{p^2+2^2 }} \right.\ \ \ \ \ \ \ \ \left \{ {{X(p)*p-4*Y(p)=\frac{16}{p^2+4} } \atop {Y(p)*p-2-X(p)=-\frac{16}{p^2+4 }} \right..$

$X(p)=Y(p)*p-2+\frac{16}{p^2+4} \ \ \ \ \Rightarrow\\(Y(p)*p-2+\frac{16}{p^2+4} )*p-4*Y(p)=\frac{16}{p^2+4} \\Y(p)*p^2-2p+\frac{16p}{p^2+4} -4*Y(p)=\frac{16}{p^2+4} \\Y(p)*(p^2-4)=2p+\frac{16}{p^2+4} -\frac{16p}{p^2+4} \\Y(p)=\frac{2p}{p^2-4}+\frac{16}{(p^2+4)*(p^2-4)} -\frac{16p}{(p^2+4)*(p^2-4)} =\\=2*\frac{p}{p^2-4}+4*\frac{2}{p^2+4}*\frac{2}{p^2-4}- 8*\frac{2}{p^2+4} *\frac{p}{p^2-4} .\\ \boxed {y(t)=2ch(2t)-4*sin(2t)*sh(2t)-8*sin(2t)*ch(2t)}.$

$X(p)=4*Y(p)+\frac{4}{p^2+4}=4*(\frac{2p}{p^2-4} +\frac{16}{(p^2+4)*(p^2-4)}-\frac{16p}{(p^2+4)*(p^2-4)} )+\frac{4}{p^2+4}=\\ =8*\frac{p}{p^2-4} +16*\frac{2}{p^2+4}*\frac{2}{p^2-4}-32*\frac{2}{p^2+4}*\frac{p}{p^2-4} +2*\frac{2}{p^2+4} .\\ \boxed {x(t)= 8*ch(2t)+16*sin(2t)*sh(2t) -32*sin(2t)*ch(2t)+2*sin(2t)}.$