I don't know why i got same x and y.

[math]x = e^{-t}(a+bt) \implies \dfrac{dx}{dt} = e^{-t}[b(1-t) - a][/math]
 
You wrote that you are substituting (equation 3) into 2*(equation 1)
The next line does not seem to reflect that. If I am wrong you must have skip some steps, which is fine by me. If you think you are correct and since I do not see these missing steps can you supply them to me.
 
\(\displaystyle \frac{dx}{dt}= -3x+ 2y\)
\(\displaystyle \frac{dy}{dt}= -2x+ y\)

Differentiate the first equation again
\(\displaystyle \frac{d^2x}{dt^2}= -3\frac{dx}{dt}+ 2\frac{dy}{dt}\)

Replace \(\displaystyle \frac{dy}{dt}\) in that using the second equation
\(\displaystyle \frac{d^2x}{dt^2}= -3\frac{dx}{dt}+ 2(-2x+ y)= -3\frac{dx}{dt}- 4x+ 2y\)

From the first equation, \(\displaystyle 2y= \frac{dx}{dt}+ 3x\)

Replacing 2y in the last equation by that
\(\displaystyle \frac{d^2x}{dt^2}= -3\frac{dx}{dt}- 4x+ \frac{dx}{dt}+ 3x\)
\(\displaystyle \frac{d^2x}{dt^2}= -2\frac{dx}{dy}- x\)
\(\displaystyle \frac{d^2x}{dt^2}+ 2\frac{dx}{dt}+ x= 0\)

That "second order linear equation with constant coefficients" in the single variable, x, has characteristic equation \(\displaystyle r^2+ 2r+ 1= (r+ 1)^2= 0\) so has a double characteristic root -1. Two independent solutions are \(\displaystyle e^{-t}\) and \(\displaystyle te^{-t}\).
The general solution, for x, is \(\displaystyle x(t)= Ae^{-t}+ Bte^{-t}\) where A and B are undetermined constants.

To find y, use
\(\displaystyle 2y= \frac{dx}{dt}+ 3x= -Ae^{-t}+ Be^{-t}- Bte^{-t}+ 3Ae^{-t}+ 3Bte^{-t}= 2Ae^{-t}+ 4Be^{-t}+ 2Bte^{-t}\)
so
\(\displaystyle y(t)= (A+ 2B)e^{-t}+ Bte^{-t}\)
 
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