Integration of a first order ODE Integrate the ODE :  $$y^\prime=2e^{5t}-7y  ,\quad y(0)=5$$
Can someone please remind me how to integrate this ODE ?
Thanks
 A: First solve the homogeneous equation $y'=-7y$ to get $y(t)=Ae^{-7t}$.
Then solve the non-homogeneous part by guessing $y(t)=ce^{5t}$ and solving for $c$ to get $c=1/6$
Your solution is now $y(t) = Ae^{-7t} + \frac{1}{6} e^{5t}$ and you use the initial condition $y(0)=5$ to solve for the constant $A$.
A: So for our differential equation we have, 
\[ y'=2e^{5t}-7y~;~~y(0)=5. \]
By inspection, we can see that this a first order linear constant coefficient non-homogeneous ordinary differential equation. The way to go about this in the less painful manner would be to use the superposition technique by combining the related homogeneous solution with the related complementary (or particular) solution and also use method of undetermined coefficients to solve for our unknown coefficients in our particular solution.
Solution.
\[ {\text{Non-homogeneous $1^{st}$ order differential}} \]
\[ y'+p(t)\cdot y = r(t) \]
So for our case here we have the following: \[ y'=2e^{5t}-7y \]
Let's first solve for the homogeneous part, $\Big({\text{i.e., setting the right hand side equal to zero}} \Big)$.
So for starts, try using ${\underline{y = e^{rt}}}$ for our problem: What do we see?
Well, rearranging and setting the RHS equal to zero we see that,
\[ ~~~y'+7y=0 \]
\[ \Rightarrow ~~~(e^{rt})'+7(e^{rt})=0 \]
So now we see that we must take one derivative of the exponential term that has a prime connected to it. So doing this leads us to:
\[ y(t)=e^{rt} \]
\[ y'(t)=re^{rt} \]
So now substituting the derivatives of y into the original ODE we can see that we get the following:
\[ re^{rt}+7(e^{rt})=0~. \]
Now let's factor out what we see in common in this new equation, which is $e^{rt}:$  
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~e^{rt}\left(r+7\right)=0~. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$ (1)
To further simplify the equation we can divide out equation $(1)$ entirely by $e^{rt}$ which would just leave us with our Characteristic Polynomial being:
\[ r+7=0~. \]
We therefore have one real-distinct root being:
 Characteristic Root is:  $r_1=-7.$
So our homogeneous solution will be the following of our characteristic root $r_1$:
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~~~~$ $y_h(t)=C_{1}e^{-7t}$.
Now we shall seek a particular solution.   
$\underline{\text{Forcing Function Case:}} \hspace{0.5in}$ $y'+7y=\underset{\text{forcing function}}{\underbrace{2e^{5t}.}}$  
$~~~~~~~~~~$
$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ f(t)=2e^{5t} $
$
\hspace{1.5in}$ Let, 
$~~~~~~~~$ 
$\hspace{2.2in}$
$ 
\begin{array}{ll}
y_{p}(t)=Ae^{5t} \\
y_{p}'(t)=5Ae^{5t}
\end{array}
$   
Substituting derivatives into differential equation:   
$5Ae^{5t}+7\left(Ae^{5t}\right)$.  
$(12A)e^{5t}=2 \cdot e^{5t}$  
After matching coefficents on the LHS with  the respective coefficients on the RHS with the corresponding function and compute the undetermined coefficient A we get the following solution:  
$
\begin{array}{ll}
12A=~2 \\
A=~\dfrac{1}{6}.
\end{array}
$   
Making our particular solution to become, 
\[ y_{p}(t)=\frac{1}{6}e^{5t}~. \]  
So now we have enough information to put together our complete solution composed of the homogeneous part plus the particular part to get:  
\[ {\underline{\text{General Solution and Particular Solution to ODE Combined}}} \]
\[ y(t)=y^{(h)}~+~y^{(p)} \]
$\hspace{2.70in} y(t)=C_{1}e^{-7t}+\dfrac{1}{6}e^{5t}.$
$\hspace{0.45in} {\underline{\text{Initial Condition}}}$ 
$\begin{array}{lll}
y(0)=5:~ 5=C_{1}e^{-7(0)}+\dfrac{1}{6}e^{5(0)} \\
~~~~~~~~~~~~~~~~~~ 5=C_{1}+\dfrac{1}{6} \\
~~~~~~~~~~~~~~~~~~ C_{1}=\dfrac{29}{6}
\end{array}
$
\[ {\underline{\text{Particular Solution to The IVP}}} \]
$\hspace{2.6in} y_{p}(t)=\dfrac{29}{6}e^{-7t}+\dfrac{1}{6}e^{5t}. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Box$
Hope this will help at all for someone. Please let me know if there is something you need clarified.
Thanks.
A: All solutions are of the form $Cf(t)+g(t)$, where $f(t)$ is a solution of the homogeneous equation $y'=-7y$, and $g(t)$ is a particular solution of the given equation.
I am sure that you know how to find a solution of the homogeneous equation.
For a particular solution, I would try $ke^{5t}$, and find the appropriate $k$ by plugging in.
Once you have the general solution, it should be straightforward to find the $C$ that makes the general solution satisfy the initial condition.
