Question

 Find the solution of

y′′−4y′+4y=112e^(6t)

with y(0)=3 and y′(0)=2.

Answer 1

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.