The characteristic equation has the single solution r=2. So the first part of the solution is of the form Ae²ᵗ+Bte²ᵗ, where A and B are constants to be found from initial conditions.
The particular solution has the form y=ae⁶ᵗ, y'=6ae⁶ᵗ, y"=36ae⁶ᵗ.
y"-4y'+4y=36ae⁶ᵗ-24ae⁶ᵗ+4ae⁶ᵗ≡112e⁶ᵗ, 16e⁶ᵗ≡112e⁶ᵗ, so a=112/16=7.
Solution is y=Ae²ᵗ+Bte²ᵗ+7e⁶ᵗ, y'=2Ae²ᵗ+2Bte²ᵗ+Be²ᵗ+42e⁶ᵗ.
y(0)=3, so A+B+7=3; y'(0)=2, so 2A+B+42=2,
Subtract the first equation from the second:
A+35=-1, A=-36; -36+B+7=3, B=32.
Solution after applying initial conditions: y=32te²ᵗ-36e²ᵗ+7e⁶ᵗ.