Chapter 6.RP, Problem 1RP

Determine the intervals for which Theorem $1$ on page $319$ guarantees the existence of a solution in that interval .

a. $y^{\left(4\right)}-\left(\ln x\right)y^{″}+x{y}^{\prime }+2y=\cos 3x$

b. $\left(x^2-1\right)y^{‴}+\left(\sin x\right)y^{″}+\sqrt{x+4}{y}^{\prime }+e^{x}y=x^2+3$

To determine

(a)

The intervals for which Theorem $1$ guarantees the existence of solution in that interval

Answer to Problem 1RP

Solution:

$\left(0,\infty \right)$ is the interval in which the theorem guarantees unique solution of given problem.

Explanation of Solution

Given:

The given differential equation is $y^{(4)}-\left(\ln x\right)y^{″}+x{y}^{\prime }+2y=\cos 3x$

Approach:

Theorem $1$ state that:

Suppose $p\left(x\right)$, …, $p\left(x\right)$ and $g\left(x\right)$ are each continuous on an interval $\left(a,b\right)$ that contains the point $x_{\circ }$ Then, for any choice of the initial values ${\gamma }_0$, ${\gamma }_1$, …${\gamma }_{n-1}$, there exists a unique solution $y\left(x\right)$ on the whole interval $\left(a,b\right)$ to the initial value problem

$y^{\left(n\right)}\left(x\right)+p_1\left(x\right)y^{\left(n-1\right)}\left(x\right)+\cdot \cdot \cdot +p_n\left(x\right)y\left(x\right)=g\left(x\right)$

So, here we will find the interval in which $\ln \left(x\right)$, $2$, $x$, $\cos 3x$ are continuous.

Calculation:

$p_1\left(x\right)=-\ln \left(x\right)$, $p_2\left(x\right)=x$, $p_3\left(x\right)=2$ and $g\left(x\right)=\cos 3x$

Now $p_1\left(x\right)$ is continuous on interval $x>0$

$p_2\left(x\right)$ and $p_3\left(x\right)$ is continuous everywhere.

$p_4\left(x\right)$ is continuous everywhere on $ℝ$

So, $\left(0,\infty \right)$ is the common interval of all

Therefore $\left(0,\infty \right)$ is the interval in which the theorem guarantees unique solution.

Conclusion:

Hence, the interval in which the theorem guarantees unique solution of the problem is $\left(0,\infty \right)$

To determine

(b)

The intervals for which Theorem $1$ guarantees the existence of solution in that interval

Answer to Problem 1RP

Solution:

$x\in \left(-4,-1\right){\displaystyle \cup \left(-1,1\right){\displaystyle \cup \left(1,\infty \right)}}$ are the intervals in which the theorem guarantees unique solution

Explanation of Solution

Given:

The given differential equation is $\left(x^2-1\right)y^{‴}+\left(\sin x\right)y^{″}+\sqrt{x+4}{y}^{\prime }+e^{x}y=x^2+3$

Approach:

We will find the interval in which coefficient of given differential equation are continuous.

Calculation:

Simplifying the given differential equation

$\begin{array}{l}\left(x^2-1\right)y^{‴}+\left(\sin x\right)y^{″}+\sqrt{x+4}{y}^{\prime }+e^{x}y=x^2+3\\ y^{‴}+\frac{\sin x}{x^2-1}{y}^{″}+\frac{\sqrt{x+4}{y}^{\prime }}{x^2-1}+\frac{e^x}{x^2-1}=\frac{x^2+3}{x^2-1} \\ \end{array}$

Here $p_1\left(x\right)=\frac{\sin x}{x^2-1}$, $p_2\left(x\right)=\frac{\sqrt{x+4}}{x^2-1}$, $p_3\left(x\right)=\frac{e^x}{x^2-1}$, and $g\left(x\right)=\frac{x^2+3}{x^2-1}$

$p_1\left(x\right)$, $p_3\left(x\right)$ and $g\left(x\right)$ is continuous for all $x^2-1\ne 0$

$p_2\left(x\right)$ is continuous for all $x+4>0$ and $x^2-1\ne 0$

So, $x\in \left(-4,-1\right){\displaystyle \cup \left(-1,1\right){\displaystyle \cup \left(1,\infty \right)}}$

Therefore, $x\in \left(-4,-1\right){\displaystyle \cup \left(-1,1\right){\displaystyle \cup \left(1,\infty \right)}}$ are the intervals in which the theorem guarantees unique solution.

Conclusion:

Hence, the interval in which the theorem guarantees unique solution are $x\in \left(-4,-1\right){\displaystyle \cup \left(-1,1\right){\displaystyle \cup \left(1,\infty \right)}}$