# Do I have something wrong when solving $y'+2y=6$?

Solve $$y'+2y=6.$$

When I do $$y'=2(3-y)\implies\int\frac{\mathrm dy}{3-y}=2\int\mathrm dx\implies-\ln{|3-y|}=2x+c\implies3-y=ke^{-2x}\therefore y=\boxed{3-ke^{-2x}},\quad c,k\in\Bbb R.$$ It satisfies the ODE because $$2ke^{-2x}+6-2ke^{-2x}=6=6.$$ However, when I try another solution, namely first solve the homogeneous equation: $$y'+2y=0\implies\int\frac{\mathrm dy}y=-2\int\mathrm dx\implies\ln{|y|}=-2x+c\implies y=ke^{-2x},\quad c,k\in\Bbb R$$ then $$y_P=k(x)e^{-2x}$$, so then $$y'_P=k'(x)e^{-2x}-2k(x)e^{-2x}\implies k'(x)e^{-2x}-2k(x)e^{-2x}+2k(x)e^{-2x}=6\implies k'(x)=6e^{2x}\implies k(x)=3e^{2x}\implies y_P=3e^{2x}e^{-2x}=3\therefore y=y_H+y_P=\boxed{3+ke^{-2x}},$$ where this solution also satisfies $$y'+2y=6$$, because $$-2ke^{-2x}+6+2ke^{-2x}=6=6.$$ My question is, how can we express both solutions with the same expression of $$y$$? I would like both solutions to be identical, but how?

Thanks!

Both solutions are the same, $$k$$ is a real number, you can write $$3+(-k)e^{-2x}$$.
They are identical. You have no reason to expect $$k$$ from one method to be exactly the same from the other method. If you include an initial condition, you'll get the same solution in both cases.
To see both are the same: $$-\ln|3-y|=2x+A \Rightarrow \ln|3-y|=-2x-A \Rightarrow |3-y|=e^{-2x-A} \Rightarrow |3-y|=e^{-A}e^{-2x} \Rightarrow 3-y=-Ce^{-2x} \Rightarrow y=3+Ce^{-2x}.$$ Alternatively (instead of variations): $$y=y_h+y_p=Ce^{-2x}+A;\\ [Ce^{-2x}+A]'+2[Ce^{-2x}+A]=6 \Rightarrow \\ 2A=6 \Rightarrow A=3 \Rightarrow \\ y=Ce^{-2x}+3.$$
• Oh, you did $e^{-A}\to-C$ when $|3-y|\to(3-y)$ instead of $e^{-A}\to C$, that is nice! Also I like your approach $\ddot\smile$. – manooooh Jan 15 at 1:15