linearly independent solutions

Question

From Calculus by Stewart (RIP):

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1. If we consider the case of Theorem 4 for Equation (2) with P, Q and R being constant, is such case proven with what comes later in the chapter namely:

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?

If not, why?

2. Where can I find a proof for Theorem 4?

Answer 1

For (2), you can use the Wronskian. Suppose that $y_1$ and $y_2$ are as given, and let $y_3$ be a solution to the differential equation $$Py'' + Qy' + Ry = 0.$$ Let $A=\left[\matrix{y_1&y_1'&y_1''\cr y_2&y_2'&y_2''\cr y_3&y_3'&y_3''}\right]$ (the transpose of the Wronskian matrix) and $\vec v = \left[\matrix{R\cr Q\cr P\cr}\right]$. Then $$A\vec v = \left[\matrix{Ry_1+Qy_1'+Py_1''\cr Ry_2+Qy_2'+Py_2''\cr Ry_3+Qy_3'+Py_3''\cr }\right]=\left[\matrix{0\cr 0\cr 0\cr}\right],$$ because $y_1$, $y_2$, and $y_3$ are all solutions.

Since $P\not=0$, that means there is a nonzero vector $\vec v$ such that $A\vec v=\vec0$; this means $A$ cannot be invertible. Hence, $\det A=0$, which means the set $\{y_1,y_2,y_3\}$ is linearly dependent.

Now, to finish it off: There are real numbers $c_1,c_2,c_3$ such that $$c_1 y_1 + c_2 y_2 + c_3 y_3 = 0.$$ We cannot have $c_3=0$, because then $\{y_1,y_2\}$ would be linearly dependent ($c_1$ or $c_2$ would have to be nonzero); hence $$y_3 = {-c_1 \over c_3}\cdot y_1 + {-c_2 \over c_3}\cdot y_2,$$ proving the theorem.

Also (and easier to prove), any linear combination is also a solution.