ODE - Subtracting complementary function from particular integral

Question

so a question is presented in a paper that is a bit confusing and some clarification would be appreciated.

Consider the ODE

$\frac{dy}{dx}-2y=e^{\lambda x}$

1) Find the general solution where $\lambda$ is a constant s.t. $\lambda \neq 2$

2) By subtracting from the particular integral an appropriate multiple of the complementary function, obtain the limit as $\lambda \rightarrow 2$ of the general solution.

Now, 1) is easy no problem there. But what do they mean by the whole subtracting from the particular integral? I would have made an Ansatz to get the required result but that is not what the question is asking, nor does it result in a limit. Attempts at subtracting and solving aren't yielding any insights either.

Any direction as to how to approach this problem would be highly appreciated.

Many thanks

K

EDIT: After multiplying through by $v(x) = e^{\int -2 dx}$ we get

$\frac{d}{dx}\left[e^{-2x}y\right] = e^{(\lambda -2)x}$

yielding the particular integral and complementary function

$y(x) = e^{2x}\int e^{(\lambda -2)x}dx + Ae^{2x}, A-constant$

EDIT2: While I am grateful to Dr. Graubner and Mr. Kermani for their responses. They do not address the issue of subtracting from the particular integral.

Answer 1

We have the particular integral

$$e^{2x}\int e^{(\lambda-2)x}dx$$ with solution $$\frac{e^{\lambda x}}{\lambda -2}$$

Recall the complementary function $$Ae^{2x}$$

Subtracting the multiple $$\frac{1}{A(\lambda-2)}$$ of the complementary function from the particular integral yields

$$\frac{e^{\lambda x}}{\lambda -2}-\frac{e^{2x}}{\lambda -2}=\frac{e^{\lambda x}-e^{2x}}{\lambda-2}$$

Taking the limit $$\lim_{\lambda \to 2} \frac{e^{\lambda x}-e^{2x}}{\lambda-2} \qquad \text{has the form} \qquad \frac{0}{0}$$

We can therefore use L'Hôpital's rule to get

$$\lim_{\lambda \to 2} \frac{xe^{\lambda x}}{1} = xe^{2 x}$$ which is the particular solution for $\lambda = 2$

Answer 2

At first solve the equation $$y'(x)-2y(x)=0$$ then multiply by $$e^{-2x}$$ and we get $$e^{-2x}y'(x)-2e^{-2x}y(x)=e^{-2x+\Lambda x}$$ this is $$\frac{d}{dx}e^{-2x}y(x)=e^{-2x+\lambda x}$$ and then integrate $$\int\frac{d}{dx}\left(e^{-2x}y(x)\right)dx=\int e^{-2x+\lambda x}dx$$

Answer 3

$$\lim _{\lambda \to 2} e^{2x}\int e^{(\lambda -2)x}dx =xe^{2x} $$

Which is the particular solution for the case of $\lambda =2$