Why Does the Solution to My Differential Equation Include a +3?

[marchingt9]

x' =-5x-y
y' =4x-y

I got
x=ae^-3t+bte^-3t
y=-2bte^-3t+2ae^-3t+be^-3t
a=0 b=0

The answer is
x=e^(-3t+3)-te^(-3+3)
y=-e^(-3+t)+2te^(-3+3)
I don't understand where the +3 comes from

 

[Geofleur]

What are the initial conditions on x and y?

 

[HallsofIvy]

e^{-3t+ 3}= (e^3)e^{-3t}. The "+ 3" just gives a specific value to your constant "a".

 

[Ray Vickson]

x' =-5x-y
y' =4x-y

I got
x=ae^-3t+bte^-3t
y=-2bte^-3t+2ae^-3t+be^-3t
a=0 b=0

The answer is
x=e^(-3t+3)-te^(-3+3)
y=-e^(-3+t)+2te^(-3+3)
I don't understand where the +3 comes from

This is not an IVP, because you have not given any initial values. That is what the "IV" stands for.