Find that solution satisfying $ϕ(1) = 3ϕ(0)$ for $y' + 5y = 2$. Find that solution satisfying $ϕ(1) = 3ϕ(0)$ for the following second order linear ordinary differential equation:

$y' + 5y = 2$

I found the solution to be $ϕ(x) = \frac{2}{5} + ce^{5x}$.


Now how do I find a particular solution satisfying $ϕ(1) = 3ϕ(0)?$ Please help me with this. In the previous part, I was asked to find the solution satisfying $ϕ(1) = 2$ which I found to be $ϕ(x) = \frac{2}{5} + \frac{8}{5}e^5e^{-5x}$ if that helps.
 A: Just plug it in into the general form.
$$\phi(1)=\frac{2}{5}+ce^{-5}\overset{!}{=}\frac{6}{5}+3c = 3\phi(0)$$
This implies that
$$c\left(3-e^{-5}\right)=-\frac{4}{5}\implies c=-\frac{4}{5\left(3-e^{-5}\right)}$$
Plugging in, you have your solution as 
$$\phi(x)=-\frac{4}{5\left(3-e^{-5}\right)}\cdot e^{-5x} + \frac{2}{5}$$
A: Firstly it should be $$y'=2-5y\\ \frac { dy }{ 2-5y } =dx\\ \int { \frac { dy }{ 2-5y }  } =\int { dx } \\ \int { \frac { d\left( 5y-2 \right)  }{ 5y-2 } =-5\int { dx }  } \\ \ln { \left| 5y-2 \right| =-5x+C } \\ 5y-2={ Ce }^{ -5x }\\$$

$$ \phi \left( x \right) =\frac { 2 }{ 5 } +{ Ce }^{ -5x }$$

then we are given $ϕ(1) = 3ϕ(0)$ so $$\frac { 2 }{ 5 } +C{ e }^{ -5 }=\frac { 6 }{ 5 } +3C\\ C\left( { e }^{ -5 }-3 \right) =\frac { 4 }{ 5 } \\ C=\frac { 4 }{ 5 } \left( { e }^{ -5 }-3 \right) $$
A: Using integrating factor:
$$e^{5x}(y'+5y) = 2 e^{5x},$$
$$\left(ye^{5x}\right)' = 2 e^{5x},$$
$$ye^{5x} = \frac25 e^{5x}+C,$$
$$y = \phi(x) = \frac25 + Ce^{-5x}.$$
Then:
$$\phi(1) = 3\phi(0),$$
$$\frac25 + Ce^{-5} = 3\left(\frac25+C\right),$$
$$C=-\frac4{5(3-e^{-5})},$$
$$\phi(x)=\frac25-\frac{4}{5(3-e^{-5})}\,e^{-5x}$$
