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given that $$y(1) = 1, y'(1) = 0$$

I have solved the general solution for this homogenous equation as follow:

which is $$r^2 + 8r -9 = 0$$ $$ r = -9, 1$$ $$y = C_1e^t + C_2e^{-9t}$$ $$y' = C_1e^t -9C_2e^{-9}$$

Now, finding the coefficients $C_1, C_2$. I am sure using regular substitution method will give the same answer but the textbook seems to use something called "Superposition" (?) using Wronskian determinants (or something along this line, I don't quite understand it)

Screenshot from an example problem

How does this work?

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Note: You forgot a $t$ in your $y'$. –  Amzoti Mar 31 '13 at 21:09

1 Answer 1

up vote 2 down vote accepted

If you have a solution of the form:

$y(t) = c_1 y_1(t) + c_2 y_2(t), y(t_0) = y_0, y'(t_0) = y'_0$

We want to find $c_1$ and $c_2$ suct that:

$c_1 y_1(t_0) + c_2y_2(t_0) = y_0$, and

$c_1y'_1(t_0) + c_2y'_2(t_0) = y'_0$

We can write this in matrix form as:

$\begin{bmatrix} y_1(t_0) & y_2(t_0)\\ y'_1(t_0) & y'_2(t_0)\end{bmatrix} \cdot \begin{bmatrix}c_1\\c_2\end{bmatrix} = \begin{bmatrix}y_0\\y'_0\end{bmatrix}$

Since our solutions are purportedly linearly independent, we know we can solve this system.

Now, how do you solve for $c_1$ and $c_2$ for this?

You can take the inverse since these are linearly independent and then pre-multiply that by the RHS. For example, look at the denominator of $c_1$ and $c_2$ and note that that is the determinant of the matrix. You will get the solution as shown in your post above.

Now, give it go with the real numbers and using this method and make sure it gives you same the result for $c_1$ and $c_2$.

Note that this is exactly as you do with the initial conditions, it is only being written more generally using this approach, but they are obviously identical.

Does that make sense?

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yes. thank you. –  40Plot Mar 31 '13 at 21:25
    
You are very welcome! Regards –  Amzoti Mar 31 '13 at 21:48
    
Nice job, plus feedback and an accept! Waiting for Babak, again, but midnight my time, so it's getting time for me to think about sleep ;-) –  amWhy Apr 11 '13 at 5:08
    
He seems to be on a different cycle than we are most days - like all nighters our time - so maybe he is a later time zone or something. Have a great night and thanks for the support! –  Amzoti Apr 11 '13 at 5:19
    
Yes, he's about 7 hours ahead of me, I'm central time...so it's 7:40 am or so for Babak. –  amWhy Apr 11 '13 at 5:42

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