\(y'-\lambda y=0\)

\(y= Ce^{at}\)

\(\Rightarrow y'= Cae^{at}\)

Plugging the values of y and y'

\(\Rightarrow \frac{Cae^{at}}{\lambda} Ce^{at}=0\)

\(\Rightarrow Ce^{at}(a-\lambda)=0\)

\(e^{at} !=0, C!=0\)

\(\Rightarrow (a-\lambda)=0\)

\(\Rightarrow a=\lambda\)

\(y= Ce^{\lambda t}\)

Lets find the particular solution

Let \(y= b+ct\)

\(\Rightarrow y'=c\)

Plugging in equation \(y'-\lambda y= 1-\lambda t\)

\(\Rightarrow c-\lambda(b+ct)= 1-\lambda t\)

\(\Rightarrow c-\lambda b-\lambda ct= 1-\lambda t\)

\(\Rightarrow c-\lambda b=1\), \(-\lambda c= -\lambda\)

Solving the second equation

\(-\lambda c= -\lambda\)

\(\Rightarrow -\lambda c+\lambda=0\)

\(\Rightarrow \lambda(-c+1)=0\)

\(\Rightarrow -c+1=0\)

\(\Rightarrow c=1\)

Plugging in 1-st equation

\(\Rightarrow 1-\lambda b=1\)

\(\Rightarrow -\lambda b=0\)

\(\Rightarrow b=0\)

The solution to the DE is

\(\Rightarrow y= Ce^{\lambda t}+t\)

Using the initial condition: \(y(0)=0\)

\(\Rightarrow y(0)= Ce^{\lambda 0}+0=0\)

\(\Rightarrow C\times1=0\)

\(\Rightarrow C=0\)

The final solution \(\Rightarrow y=0e^{\lambda t}+t\)

\(\Rightarrow y=t\)