# Equating Coefficients - Meaning and Example

I am trying to understand what "Equating the Coefficients " means

I am given the following:

$$t_1 \left(t_2+\frac{1}{x+1}\right)+\frac{2 t_2}{x}=t_1^2 b_2'+t_1 \left(2 b_2 t_1'+b_1'\right)+b_1 t_1'+b_0'+\frac{d R}{d x}$$

and the author continues...

"Equating coefficients by $t_1$, we get the following system of equations"

$$\begin{eqnarray} b_2' & = & 0 \\ 2 b_2 t_1'+b_1' & = & t_2+\frac{1}{x+1}\\ b_1 t_1'+b_0'+\frac{d R}{d x} & = & \frac{2 t_2}{x} \end{eqnarray}$$

"From the first equation, we find"

$$b_2=c_2$$

"From the second equation, we find"

$$\begin{array}{cc} b_1+2 c_2 t_1 & =\int (t_2+\frac{1}{x+1}) \\ \end{array}$$

I am unclear how they are arriving at this result. Help would be appreciated.

• Notice the expression $t_1 \left(t_2+\frac{1}{x+1}\right)$ on the LHS and the expression $t_1 \left(2 b_2 t_1'+b_1'\right)$? Because of their similar form, we can equate those and then you equate the other similar expressions, so you can figure out how to make the LHS = RHS. Can you now figure it out? – Moo Mar 18 '17 at 13:42
• Think at your equality as an equality between polynomials in the indeterminate $t_1$ but with "parametric" coefficients. What does it mean that two polynomials are equal? – Ender Wiggins Mar 18 '17 at 13:44
• @Moo - Ok so I see that one. But why wouldn't $t_1 \left(t_2+\frac{1}{x+1}\right)$ equate with $b_1 t_1$ – PiE Mar 18 '17 at 13:48
• I find the existence of the $t_{1}'$ makes all this pretty dodgy; it seems to me to undercut the "$t_1$ is an indeterminate" approach (which I agree is what is usually intended). – ancientmathematician Mar 18 '17 at 13:49

Basically, it means equating coefficients of each linearly independent variable from LHS and RHS. So that the identity 0=0 remains true for any value of the variables.

Here the linearly independent variables are $t_1$ and $t_1^2$ .

• Ok - I think I got this. Thanks. – PiE Mar 18 '17 at 15:52