Let $f: R \rightarrow R$ be a function satisfying $f(x+2 y)=f(x) e^{2 y}+f(2 y) e^x+x^2\left(1-e^{2

Question

Let $f: R \rightarrow R$ be a function satisfying $f(x+2 y)=f(x) e^{2 y}+f(2 y) e^x+x^2\left(1-e^{2 y}\right)+4 y^2\left(1-e^x\right)+4 x y \forall x, y \in R$ and $f^{\prime}(0)=1$, then:

Professor PC Thomas Classes · Answer

$f(x+2y)=f(x)e^{2y}+f(2y)e^{x}+x^{2}(1-e^{2y})+4y^{2}(1-e^{x})+4xy$...(1) Putting $x=y=0$ in equation (1), we get $f(0)=2f(0)\Rightarrow f(0)=0$ Now, $f^{\prime}(x)=lim_{y\rightarrow0}\frac{f(x+2y)-f(x)}{2y}$ $=lim_{y\rightarrow0}\frac{f(x)e^{2y}+e^{x}f(2y)+x^{2}(1-e^{2y})+4y^{2}(1-e^{x})+4xy-f(x)}{2y}$ $=lim_{y\rightarrow0}\frac{f(x)(e^{2y}-1)+e^{x}(f(2y)-f(0))-(e^{2y}-1)x^{2}+4y^{2}(1-e^{x})+4xy}{2y}$ $=f(x)lim_{y\rightarrow0}\frac{(e^{2y}-1)}{2y}+e^{x}lim_{y\rightarrow0}\frac{f(2y)-f(0)}{2y}-x^{2}lim_{y\rightarrow0}\frac{(e^{2y}-1)}{2y}+2x+lim_{y\rightarrow0}2y(1-e^{x})$ $f^{*}(x)=f(x)+e^{x}-x^{2}+2x$ $e^{-x}f^{\prime}(x)-e^{-x}f(x)=1-x^{2}e^{-x}+2xe^{-x}$ $\frac{d}{dx}(e^{-x}f(x))=\frac{d}{dx}(x+x^{2}e^{-x})\Rightarrow e^{-x}f(x)=x+x^{2}e^{-x}+c$ put $x=0$ we get $c=0$ Hence $f(x)=xe^{x}+x^{2}$