Find the particular solution of the following equation 
my task is to Find the particular solution of the following equation at y(0) =1. the equation is
$$dy/dx + 4y = 7$$

I have made this much progress as far as separating the Ys and the Xs and taking the anti derivatives (I hope it's correct) but don't know how to reach an answer
dy/dx + 4y = 7    || .dx
$dy + 4y = 7 dx$
$4 y^2/2 = 7x +c$
.....
correct answer is:
$y = 7/4 - 3/4e^{-4x}$
update:
I managed to get 7/4 and Ce^-4x) but still don't get where -3/4 comes from
thank you!
 A: We have 
$$\frac{dy}{dx}=7-4y,$$ and therefore 
$$\frac{dy}{7-4y}=dx.$$
Continue.  
Added: Integrate. We get $-\frac{1}{4}\ln(|7-4y|)=x+C$. Put $x=0$. We get $C=-\frac{1}{4}\ln 3$. Thus
$$\ln(7-4y)=-4x+\ln 3.$$
Note this is valid only when $7-4y\gt 0$. Exponentiate. We get $7-4y=3e^{-4x}$.
Solve for $y$. We get $y=\frac{1}{4}\left(7-3e^{-4x}\right)$. 
A: $\frac{dy}{dx}+4Y=7$ |
can be solved letting $y=Ae^{bx}$.
$y'=Abe^{-x}$ substituting  into the original equation and
solving the homogeneous part  gives
$(b+4)=0 .$
hence $b=-4 $ we have a solution $y=Ae^{-4x}$
called the complementary solution ,we need o find the particular solution
we check the right side of the equation ,the guess for a constant is $C$
$$Y(particular)=c$$
$$y'(particular)=0$$
substituting back into $\frac{dy}{dx}+4y=7$
we have $$4c=7$$
$$c=7/4$$
applying superposition whereby we have the general sol as $$y=Ae^{-4x}+7/4$$ the sum of complementary solution and particular solution .$y(0)=1$
$$1=A+7/4  \phantom{filler}\text{  (ANY NUMBER TO EXPONENT ZERO IS ONE)}$$
$$A =-3/4$$
$$Y=-3/4(e^{-4x})+7/4.$$
