Differential equation first order what have I done wrong?

Question

\begin{align} & f'(x) + 4f(x) = 6e^{-2x}; \quad f(0) = 1 \\ \implies & s \mathscr{F}(s) - f(0) + 4 \mathscr{F}(s) = \dfrac{6}{s+2} \\ \implies & \mathscr{F}(s)(s+4)= \dfrac{6}{s+2} - 1 \\ \implies & \mathscr{F}(s)(s+4)= \dfrac{6-s-2}{s+2} \\ \implies & \mathscr{F}(s)= \dfrac{4-s}{(s+2)(s+4)} = \dfrac{A}{s+2} - \dfrac{B}{s+4} \end{align}

Using partial fraction, we have \begin{align} & 4-s = A(s+4) - B(s+2) \\ \implies & 4 (-2+4)-(-1)(-4+2) = 4-2 \\ \implies & \dfrac{4}{s+2} - \dfrac{2}{s+4} \rightarrow (4e^{-2x}) - (2e^{-4x}) \end{align}

But it should be $e^{-4x}(3e^{2x} - 2$)

Answer 1

The given eqn:$$\ f'(x)+4f(x) = 6e^{-2x} ; f(0) = 1$$ Or, it can be written as:$$\ \frac{dy}{dx}+4y = 6e^{-2x}$$This is linear differential equation in $\ y$ and hence its integrating factor(I.F): $$\ e^{\int 4dx} = e^{4x}$$ Hence solution is : $$\ y(e^{4x}) = 6\int e^{-2x}.e^{4x}dx + c$$$$\ y(e^{4x}) = 6\int e^{2x}dx + c$$$$\ y(e^{4x}) = 6(\frac{e^{2x}}{2}) + c$$$$\ y(e^{4x}) = 3e^{2x} + c$$Now , when $\ x = 0, y = 1$.Therefore putting these values in last eqn:$$\ 1(e^0) = 3e^0 + c$$$$\ c = (-2)$$ Therefore the solution is:$$\ y = 3e^{-2x} - 2e^{-4x}$$$$\ f(x) = 3e^{-2x} - 2e^{-4x}$$

Answer 2

Here you made a sign mistake

$$ \mathscr{F}(s)(s+4)= \dfrac{6}{s+2} \color {red}{- 1} $$

Should be $+f(0)=+1$

Then

$$ \mathscr{F}(s)= \dfrac{(s+8)}{(s+4)(s+2)} =\dfrac{A}{(s+2)}+\dfrac{B}{(s+4)} \to A=3, B=-2$$ The solution is $$f(x)=3e^{-2x}-2e^{-4x}$$