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

x=e−t(a+bt)  ⟹  dxdt=e−t[b(1−t)−a]x = e^{-t}(a+bt) \implies \dfrac{dx}{dt} = e^{-t}[b(1-t) - a]
 
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.
 
dxdt=−3x+2y\displaystyle \frac{dx}{dt}= -3x+ 2y
dydt=−2x+y\displaystyle \frac{dy}{dt}= -2x+ y

Differentiate the first equation again
d2xdt2=−3dxdt+2dydt\displaystyle \frac{d^2x}{dt^2}= -3\frac{dx}{dt}+ 2\frac{dy}{dt}

Replace dydt\displaystyle \frac{dy}{dt} in that using the second equation
d2xdt2=−3dxdt+2(−2x+y)=−3dxdt−4x+2y\displaystyle \frac{d^2x}{dt^2}= -3\frac{dx}{dt}+ 2(-2x+ y)= -3\frac{dx}{dt}- 4x+ 2y

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

Replacing 2y in the last equation by that
d2xdt2=−3dxdt−4x+dxdt+3x\displaystyle \frac{d^2x}{dt^2}= -3\frac{dx}{dt}- 4x+ \frac{dx}{dt}+ 3x
d2xdt2=−2dxdy−x\displaystyle \frac{d^2x}{dt^2}= -2\frac{dx}{dy}- x
d2xdt2+2dxdt+x=0\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 r2+2r+1=(r+1)2=0\displaystyle r^2+ 2r+ 1= (r+ 1)^2= 0 so has a double characteristic root -1. Two independent solutions are e−t\displaystyle e^{-t} and te−t\displaystyle te^{-t}.
The general solution, for x, is x(t)=Ae−t+Bte−t\displaystyle x(t)= Ae^{-t}+ Bte^{-t} where A and B are undetermined constants.

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