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#### karush

##### Well-known member

ok I did this first

$u(t)y'+u(t)y=u(i)5\sin{2t}$

then

$\frac{1}{5}u(t)y'+\frac{1}{5}u(t)y=u(i)\sin{2t}$

so far .... couldn't find an esample to follow

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book answer

$y=ce^{−t}+\sin 2t−2\cos 2t$

$\textit{y is asymptotic to sin2t−2cos2t as $t\to\infty$}$