# Differential equations and Cramer's rule

1. Apr 16, 2014

### JasonHathaway

Hi everyone,

I'm taking the Differential Equations for the first time, and I want to know the most helpful textbook for the subject.

Find the differential equation which its general solution is:
$y=C_{1}+C_{2} x+x^{2}$

Solution:
$y^{'}=0+C x+2x$
$y^{''}=0+0+2$

Solving:

$\begin{bmatrix} y-x^{2} & 1 & x \\ y^{'}-2x & 0 & 1 \\ y^{''}-2 & 0 & 1 \end{bmatrix} y^{''}=2$

How and why did the equations become like that?

Last edited: Apr 16, 2014
2. Apr 17, 2014

### Staff: Mentor

The above should be y' = 0 + C2 + 2x
You have these equations:
y = C1 + C2x + x2
y' = C2 + 2x
y'' = 2

The three equations above can be rewritten as
y - x2 = C1 + C2x
y' - 2x = C2
y'' - 2 = 2

Now why someone would want to do this, I have no idea, nor do I understand what the matrix means. Also, as you wrote it, you have the matrix times y'' being equal to 2. I don't know if that's what you meant to write.

Generally, when you're working with differential equations, you are given an equation, and asked to find the solution, not the other way around, as in this problem.

Since you know the solution (y = C1 + C2x + x2), and have determined from it that y'' = 2, that seems to me to be all you need to do. That's your differential equation, and its general solution (which you can get by integrating it twice) would be y = C1 + C2x + x2.

3. Apr 18, 2014

### Bill_K

You have three linear equations for two unknowns, C1 and C2. Cramer's Rule gives the consistency condition, namely that the determinant of the coefficients is zero. In this case,
$$\left|\begin{array}{ccc}1&x&y - x^2\\0&1&y' - 2x\\0&0&y'' - 2\end{array}\right| = 0$$
or y'' = 2.