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## Linear Homogeneous ODEs and Wronskian

Consider the linear homogeneous second order ODE $y''+p(x)y'+q(x)y=0. \;\;\;\;(31)$ Principle of superposition: If $$y_1$$ and $$y_2$$ are solutions of (31), then $$c_1y_1+c_2y_2$$ is also a solution of (31). (verify)

Question. If $$y$$ is a solution of (31), is it true that $$y=c_1y_1+c_2y_2$$ for some scalars $$c_1$$ and $$c_2$$?
Answer. Yes if $$y_1$$ and $$y_2$$ are linearly independent functions.

Definition. $$y_1,y_2,\ldots,y_n$$ are linearly independent functions if $c_1y_1+c_2y_2+\cdots+c_ny_n=0\implies c_1=c_2=\cdots=c_n=0.$
Theorem. $$y_1,y_2,\ldots,y_n$$ are linearly dependent functions if and only if at least one of them is a linear combination of others.

Example.

1. $$y_1=x+1$$ and $$y_2=x-1$$ are linearly independent. Because if $$c_1y_1+c_2y_2=0$$, then \begin{align*} &c_1(x+1)+c_2(x-1)=0\\ \implies& (c_1+c_2)x+(c_1-c_2)=0\\ \implies& c_1+c_2=0,\; c_1-c_2=0\\ \implies& c_1=c_2=0. \end{align*}
2. $$y_1=e^x$$ and $$y_2=e^{2x}$$ are linearly independent.

3. $$y_1=x$$ and $$y_2=2x$$ are linearly dependent since $$-2y_1+y_2=0$$.

The Wronskian of $$y_1$$ and $$y_2$$ is denoted by $$W(y_1,\,y_2)$$ and defined by $W(y_1,\,y_2)=\begin{array}{|cc|}y_1&y_2\\y_1'&y_2'\end{array}=y_1y_2'-y_2y_1'.$
Example.

1. $$W(x+1,\,x-1)=\,\begin{array}{|cc|}x+1&x-1\\(x+1)'&(x-1)'\end{array}=\,\begin{array}{|cc|}x+1&x-1\\1&1\end{array}=(x+1)\cdot 1-(x-1)\cdot 1=2\neq 0.$$

2. $$W(e^x,\,e^{2x})=\,\begin{array}{|cc|} e^x&e^{2x}\\(e^x)'&(e^{2x})'\end{array}=\,\begin{array}{|cc|}e^x&e^{2x}\\e^x&2e^{2x}\end{array}=e^x\cdot 2e^{2x}-e^{2x}\cdot e^x=e^{3x}\neq 0.$$

Theorem. $$y_1$$ and $$y_2$$ are linearly independent functions if and only if $$W(y_1,\,y_2)\neq 0$$. In other words, $$y_1$$ and $$y_2$$ are linearly dependent functions if and only if $$W(y_1,\,y_2)=0$$.

$$(\Longrightarrow)$$ Assume that $$y_1$$ and $$y_2$$ are linearly dependent functions. Then one of $$y_1$$ and $$y_2$$ is a multiple of the other, say $$y_2=cy_1$$ for some scalar $$c$$. Then $W(y_1,\,y_2)=W(y_1,\,cy_1)=y_1\cdot c_1y_1'-cy_1\cdot y_1'=0.$ $$(\Longleftarrow)$$ Assume that $$W(y_1,\,y_2)=y_1y_2'-y_2y_1'=0$$. Note that $\frac{d}{dx}\left(\frac{y_2}{y_1} \right)=\frac{y_1y_2'-y_2y_1'}{y_1^2}=0.$ Then $$\frac{y_2}{y_1}=c,$$ constant. Thus $$y_2=cy_1$$ and consequently $$y_1$$ and $$y_2$$ are linearly dependent functions. $$\blacksquare$$

Theorem. Suppose that $$y_1$$ and $$y_2$$ are solutions of the IVP $y''+p(x)y'+q(x)y=0,\;\; y(x_0)=r,\, y'(x_0)=s.$ Then $$c_1y_1+c_2y_2$$ is also a solution for some scalars $$c_1$$ and $$c_2$$ if and only if $$W(y_1,\,y_2)(x_0)\neq 0$$.

$$(\Longleftarrow)$$ Assume that $$W(y_1,\,y_2)(x_0)\neq 0$$. We will find $$c_1$$ and $$c_2$$ such that $$c_1y_1+c_2y_2$$ is also a solution of the given IVP. First note that since $$y_1$$ and $$y_2$$ are solutions of $$y''+p(x)y'+q(x)y=0$$, $$c_1y_1+c_2y_2$$ is also a solution of $$y''+p(x)y'+q(x)y=0$$ by the principle of superposition. Now we will find $$c_1$$ and $$c_2$$ such that $$c_1y_1+c_2y_2$$ satisfies the initial conditions. \begin{align*} y(x_0)=r&\implies c_1y_1(x_0)+c_2y_2(x_0)=r\\ y(x_0)=s&\implies c_1y_1'(x_0)+c_2y_2'(x_0)=s \end{align*} Since $$W(y_1,\,y_2)(x_0)\neq 0$$, solving for $$c_1$$ and $$c_2$$ we get $c_1=\frac{ry_2'(x_0)-sy_2(x_0)}{y_1(x_0)y_2'(x_0)-y_2(x_0)y_1'(x_0)}=\frac{ry_2'(x_0)-sy_2(x_0)}{W(y_1,\,y_2)(x_0)},$ $c_2=\frac{-ry_1'(x_0)+sy_1(x_0)}{y_1(x_0)y_2'(x_0)-y_2(x_0)y_1'(x_0)}=\frac{-ry_1'(x_0)+sy_1(x_0)}{W(y_1,\,y_2)(x_0)}.$ $$(\Longrightarrow)$$ Similar. $$\blacksquare$$

Theorem. Suppose that $$y_1$$ and $$y_2$$ are solutions of $y''+p(x)y'+q(x)y=0.$ Then any solution $$y$$ can be written as $$y=c_1y_1+c_2y_2$$ for some scalars $$c_1$$ and $$c_2$$ if and only if $$W(y_1,\,y_2)(x_0)\neq 0$$ for some $$x_0$$.

$$(\Longleftarrow)$$ Assume $$W(y_1,\,y_2)(x_0)\neq 0$$. Consider an arbitrary solution $$y=\phi(x)$$ of the IVP $y''+p(x)y'+q(x)y=0,\; y(x_0)=\phi(x_0),y'(x_0)=\phi'(x_0).$ By the preceding theorem $$y=c_1y_1+c_2y_2$$ is a solution for some scalars $$c_1$$ and $$c_2$$. Then $$\phi=y=c_1y_1+c_2y_2$$ by the uniqueness of the solution of the preceding IVP.

$$(\Longrightarrow)$$ It follows from the linear independence of the fundamental solutions $$y_1$$ and $$y_2$$. $$\blacksquare$$

Note:

1. It can be shown for $$y_1$$ and $$y_2$$ in the preceding theorem that $$W(y_1,\,y_2)(x)=0$$ either for all values of $$x$$ or no values of $$x$$. (see Abel's theorem)

2. When $$W(y_1,\,y_2)\neq 0$$, $$y_1$$ and $$y_2$$ are linearly independent and any solution $$y$$ can be written as $$y=c_1y_1+c_2y_2$$ for some scalars $$c_1$$ and $$c_2$$. That is why they are called fundamental solutions of $$y''+p(x)y'+q(x)y=0$$ and the general solution of $$y''+p(x)y'+q(x)y=0$$ is $y=c_1y_1+c_2y_2$ for arbitrary scalars $$c_1$$ and $$c_2$$.

Example. Recall from Section 3.1 that $$y''-2y'-3y=0$$ has the general solution $y=c_1e^{-x}+c_2e^{3x}.$ Here $$y_1=e^{-x}$$ and $$y_2=e^{3x}$$ are fundamental solutions because $W(e^{-x},\,e^{3x})=\,\begin{array}{|cc|}e^{-x}&e^{3x}\$$e^{-x})'&(e^{3x})'\end{array} =\,\begin{array}{|cc|}e^{-x}&e^{3x}\\-e^{-x}&3e^{3x}\end{array} =e^{-x}\cdot 3e^{3x}-e^{3x}\cdot(-e^{-x})=4e^{2x}\neq 0.$ Example. Consider \(y_1=e^{r_1x}$$ and $$y_2=e^{r_2x}$$, $$r_1\neq r_2$$. $W(e^{r_1x},\,e^{r_2x})=\,\begin{array}{|cc|}e^{r_1x}&e^{r_2x}\$$e^{r_1x})'&(e^{r_2x})'\end{array} =\,\begin{array}{|cc|}e^{r_1x}&e^{r_2x}\\r_1e^{r_1x}&r_2e^{r_2x}\end{array} =e^{r_1x}\cdot r_2e^{r_2x}-e^{r_2x}\cdot r_1e^{r_1x})=(r_2-r_1)e^{(r_1+r_2)x}\neq 0.$ For \(ay''+by'+cy=0$$, if the roots $$r_1$$ and $$r_2$$ of the characteristic equation $$ar^2+br+c=0$$ are real and distinct, then $$y=e^{r_1x}$$ and $$y=e^{r_2x}$$ are fundamental solutions and the general solution is $y=c_1e^{r_1x}+c_2e^{r_2x}.$

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