Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Integrate the ODE : $$y^\prime=2e^{5t}-7y ,\quad y(0)=5$$

Can someone please remind me how to integrate this ODE ?

Thanks

share|improve this question
1  
Thanks Raeder, I was about to make the same edit! –  Chris Taylor May 10 '11 at 15:12
1  
All caps is interpreted as shouting. Did you really mean to yell "First order"? –  Arturo Magidin May 10 '11 at 15:13

3 Answers 3

up vote 4 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
    
Thank you very much, but the other reply gave a more direct answer –  NLed May 10 '11 at 15:24

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 $A$.

share|improve this answer
    
Thanks, thats all i needed... ! –  NLed May 10 '11 at 15:23
    
just from looking at it, it is obvious that $y(t)=ce^{5t}$ is a solution. But, what I've always struggled with is, is there a systematic way of choosing the particular solution? My issue is it seems odd that we can systematically, for the most part, find the homogenous solution, yet in finding the particular solution, we are reduced to guessing. Yes, they're educated guesses, but there is still an element trial and error that is bothersome to me. –  rcollyer Jun 1 '11 at 17:26
1  
@rcollyer There are a few good heuristics for finding a particular solution, but I'm not aware of a general method that works in all cases. The method you've seen is called the "Method of unknown coefficients", aka the "lucky guess" method, check it here. There's a nice table of good guesses given the form of the inhomogeneity. Sometimes the method of variation of parameters or the annihilator method are useful. –  Chris Taylor Jun 1 '11 at 18:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.