Answer: \(\lambda=2\).

The solution is \(y=3e^{-x}-e^{-4x}+2xe^{-x}\).

Let the particular integral be \(y_p=\lambda xe^{-x}\).

Differentiate:

\(y_p' = \lambda e^{-x}-\lambda xe^{-x}\)

\(y_p'' = -2\lambda e^{-x}+\lambda xe^{-x}\)

Substitute into \(y''+5y'+4y=6e^{-x}\):

\(\begin{aligned}y_p''+5y_p'+4y_p &= \left(-2\lambda e^{-x}+\lambda xe^{-x}\right)+5\left(\lambda e^{-x}-\lambda xe^{-x}\right)+4\lambda xe^{-x}\\ &= 3\lambda e^{-x}.\end{aligned}\)

So \(3\lambda e^{-x}=6e^{-x}\), hence \(\lambda=2\).

Now solve the differential equation. The complementary equation is

\(m^2+5m+4=0\), so \((m+1)(m+4)=0\).

Therefore the complementary function is \(y_c=Ae^{-x}+Be^{-4x}\).

Using the particular integral found above, the general solution is

\(y=Ae^{-x}+Be^{-4x}+2xe^{-x}.\)

Differentiate:

\(y'=-Ae^{-x}-4Be^{-4x}+2e^{-x}-2xe^{-x}.\)

Use the initial conditions at \(x=0\):

From \(y(0)=2\), \(A+B=2\).

From \(y'(0)=3\), \(-A-4B+2=3\), so \(2-A-4B=3\), giving \(A+4B=-1\).

Solve the simultaneous equations:

\(A+B=2\), \(A+4B=-1\).

Subtracting gives \(3B=-3\), so \(B=-1\), and then \(A=3\).

Hence the required solution is \(y=3e^{-x}-e^{-4x}+2xe^{-x}\).