Find the function ' $f$ ' which satisfies the equation $\dfrac{d f}{d x}=2 f$, given that $f(0)=e^3$

The given differential equation is $\dfrac{df}{dx} = 2f$.

Separating the variables, we have $\dfrac{df}{f} = 2 dx$.

Integrating both sides, $\int \dfrac{df}{f} = \int 2 dx$, which gives $\ln|f| = 2x + C$.

Exponentiating both sides, $f(x) = e^{2x + C} = Ae^{2x}$, where $A = e^C$.

Using the initial condition $f(0) = e^3$, we substitute $x = 0$ into the equation: $e^3 = Ae^{2(0)} = A(1)$, so $A = e^3$.

Substituting the value of $A$ back into the expression for $f(x)$, we get $f(x) = e^3 \times e^{2x} = e^{2x + 3}$.

Answer: $e^{2 x+3}$