"\\frac{d^2y}{dt^2}+2\\frac{dy}{dt}-3y=\\sin t"
The corresponding homogeneous equation "y''+2y'-3y=0" has the characteristics equation "m^2+2m-3=0"
"(m-1)(m+3)=0\\\\\nm=1\\text{ or } m=-3"
"Y_c=C_1e^{1.t}+C_2e^{-3t}\\\\\nY_c=C_1e^t+C_2e^{-3t}"
The non-homogeneous term is "g(t)=\\sin t" . We choose "Y_p=A\\cos t+ B\\sin t"
"Y'_p=B\\cos t-A\\sin t, Y''_p=-(A\\cos t+B\\sin t)" . Put these into the equation. We have;
"-(A\\cos t+B\\sin t)+2(B\\cos t-A\\sin t)-3(A\\cos t+B\\sin t)=\\sin t\\\\\n (-4A+2B)\\cos t+(-2A-4B)\\sin t=\\sin t"
Compare both sides
"-4A+2B=0.................(1)\\\\\n-2A-4B=1................(2)"
From (1), "A=\\frac{B}{2}." Put this into (2)
"-5B=1,B=-\\frac{1}{5}\\\\\nA=-\\frac{1}{10}"
Therefore,"Y_p=-\\frac{1}{10}\\cos t-\\frac{1}{5}\\sin t"
The general solution "y=Y_c+Y_p"
"y=C_1e^t+C_2e^{-3t}-\\frac{1}{10}\\cos t-\\frac{1}{5}\\sin t"
But, "y=y'=0" when "t=0" . This implies that,
"0=C_1e^0+C_2e^{-3.0}-\\frac{1}{10}\\cos 0-\\frac{1}{5}\\sin 0\\\\\n0=C_1+C_2-\\frac{1}{10}\\\\\n\\frac{1}{10}=C_1+C_2....................(3)"
"y'=C_1e^t-2C_3e^{-3t}+\\frac{1}{10}\\sin t-\\frac{1}{5}\\cos t\\\\\n0=C_1e^0-3C_2e^{-3.0}+\\frac{1}{10}\\sin 0-\\frac{1}{5}\\sin 0\\\\\n0=C_1-3C_2-\\frac{1}{5}\\\\\nC_1-3C_2=\\frac{1}{5}....................(4)"
Subtract (4) from (3)
"4C_2=\\frac{1}{10}-\\frac{1}{5}=-\\frac{1}{10},C_2=-\\frac{1}{40}"
Put this into (3)
"C_1=\\frac{1}{10}+\\frac{1}{40}=\\frac{1}{8}"
Hence, "y=\\frac{1}{8}e^t-\\frac{1}{40}e^{-3t}-\\frac{1}{10}\\cos t-\\frac{1}{5}\\sin t"
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