$If\;\phi \left( x \right) = \;\phi '\left( x \right)\;and\;\;\phi \;\left( 1 \right) = 2,\;then\;\;\phi \left( 3 \right)\;equals\;$

Let,

$\begin{array}{c} y=\phi(x) \\ \therefore \frac{d y}{d x}=\phi^{\prime}(x) \\ \phi(x)=\phi^{\prime}(x) \end{array}$

$\begin{array}{l}\Rightarrow y=\frac{d y}{d x} \\ \Rightarrow \frac{d y}{d x}=y \\ \Rightarrow \frac{d y}{y}=d x \\ \Rightarrow \int \frac{d y}{y}=\int x \\ \Rightarrow \ln |y|=x+\ln c\end{array}$

$\begin{array}{l}\Rightarrow \ln |y|-\ln e=x \\ \Rightarrow \ln \left(\frac{y}{c}\right)=x\end{array}$

$\begin{array}{l}\Rightarrow \ln (k y)=x \\ \Rightarrow k y=e^{x} \\ \Rightarrow y=k_{1} e^{x}\end{array}$

$\frac{1}{e}=k=$ constant $=k_{1}$

$\begin{array}{l}y=k_{1} e^{x} \\ \Rightarrow \phi(x)=k_{1} e^{x} \\ \Rightarrow \phi(1)=k_{1} e \\ \Rightarrow 2=k_{1} e \\ \Rightarrow k_{1}=\frac{2}{e}\end{array}$

$\begin{array}{l}\therefore \phi(x)=\frac{2}{e} \cdot e^{x} \\ \Rightarrow \phi(x)=2 e^{(x-1)}\end{array}$

$\Rightarrow \phi(3)=2 \cdot e^{(3-1)}=2 e^{2}$