_{h}(t)=c

_{1}e

^{-t}(1,3)+c

_{2}e

^{t}(1,1)

_{1}and c

_{2}are functions of t. Plug this back into the original equation. You will get:

_{1}'(t)e

^{-t}(1,3)+c

_{1}(t)(e

^{-t})'(1,3)+c

_{2}(t)'e

^{t}(1,1)+c

_{2}(t)(e

^{t})'(1,1)={(2,3);(-1,-2)}[c

_{1}(t)e

^{-t}(1,3)+c

_{2}(t)e

^{t}(1,1)]+(e

^{t},t);

_{1}'(t)e

^{-t}(1,3)+c

_{1}(t)(e

^{-t})'(1,3)+c

_{2}(t)'e

^{t}(1,1)+c

_{2}(t)(e

^{t})'(1,1)=-c

_{1}(t)e

^{-t}(1,3)+c

_{2}(t)e

^{t}(1,1)+(e

^{t},t)

^{-t}an e

^{t}on the left side in the second and fourth terms, to obtain:

_{1}'(t)e

^{-t}(1,3)-c

_{1}(t)e

^{-t}(1,3)+c

_{2}(t)'e

^{t}(1,1)+c

_{2}(t)e

^{t}(1,1)=

_{1}(t)e

^{-t}(1,3)+c

_{2}(t)e

^{t}(1,1)+(e

^{t},t)

_{1}(t)'(1,3)e

^{-t}+c

_{2}(t)'(1,1)e

^{t}=(e

^{t},t)

_{1}' and c

_{2}'. After solving it we obtain:

_{1}'(t)=(te

^{t}-e

^{2t})/2;

_{2}'(t)=(3-te

^{-t})/2;

_{1}(t)=(t-1)e

^{t}/2-e

^{2t}/4+S

_{1};

_{2}(t)=[3t+(t+1)e

^{-t}]/2+S

_{2};

^{ }

_{inh}=S

_{1}e

^{-t}(1,3)+S

_{2}e

^{t}(1,1)+(2t-2-e

^{t})/4*(1,3)+(3te

^{t}+t+1)/2*(1,1)

_{inh}=S

_{1}e

^{-t}(1,3)+S

_{2}e

^{t}(1,1)+(t-e

^{t}/4+(3/2)te

^{t},2t-¾e

^{t}+(3/2)te

^{t}-1)

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