Lesson 12 - Theory of Linear Differential Equations (Theorem 1)

Consider the $n$th order linear differential equation.

$y^{(n)}(x)+p_1(x)y^{(n-1)}(x)+p_2(x)y^{(n-2)}(x)\cdots +p_{(n-1)}{y}^{\prime }(x)+p_{n}y(x)=f(x)$

with the initial conditions

$y(x_0)=a_o$, $y^{\prime }(0)=a_1$,$\dots$, $y^{(n-1)}(x_0)=a_{(n-1)}$

and where $p_1(x)$, $p_2(x)$,$\dots$, $p_n(x)$ are functions of $x$ alone.

Theorem 1

Existence and Uniqueness: Suppose $p_1(x)$,$\dots$,$p_n(x)$ and $f(x)$ are each continuous on an interval $I=(a,b)$ that contains the point $x_0$. Then for any choice of the initial values $a_0,a_1,\dots ,a_{(n-1)}$, there exists a unique solution $y(x)$ on the whole interval $I$ to above initial value (IVP).

Example

For the IVP determine the values of $x_0$ and the intervals $(a,b)$ that contains $x_0$ for which the above existence and uniqueness theorem guarantees the existence of a unique solution on $(a,b)$.

$x(x-1)y^{‴}-4xy-6x^3{y}^{\prime }-y\sin x=\sqrt{x+3}$, $y(x_0)=1,y^{\prime }(x_0)=0,y^{″}(x_0)=7$

First, this equation needs to be put into standard form. We divide by the leading coefficient $x(x-1)$.

We combine the two $y$ terms.

To use theorem 1, we need $f(x)$, $p_1(x)$ and $p_2(x)$ to be continuous.

$p_{1(x)}=\frac{6x^2}{x-1}$, $p_2(x)=\frac{4x+\sin x}{x(x-1)}$, and $f(x)=\frac{\sqrt{x+3}}{x(x+1)}$

For $p_{1(x)}$, $x$ cannot be $1$.

For $p_2(x)$, $x$ cannot be $0$ or $1$.

For $f(x)$, $x$ must be greater than or equal to $-3$, and $x$ cannot be $0$ or $1$.

So the intervals that contain $x_0$ are $[-3,0)\cup (0,1)\cup (1,\mathrm{\infty })$.

The next lesson will be a mini-lesson on the differential operator $L$. As such, the following lessons will be sub-lessons of this lesson and will be labeled $12.1$, $12.2$, and so-on.

Next Lesson: Lesson 12.1 - The Differential Operator

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