Q Review Problems-1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the intervals for which Theorem guarantees the existence of a solution in that

(a)  ${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{-}\mathbf{\left(}\mathit{l}\mathit{n}\mathit{x}\mathbf{\right)}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{3}\mathit{x}$

(b) $\left(x^2-1\right){\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{\left(}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{\right)}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\sqrt{\mathbf{x}\mathbf{+}\mathbf{4}}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}{\mathit{e}}^{\mathbf{x}}\mathit{y}\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{ }$

Step-by-Step Solution

Answer

1.  $(0,\infty )$ is interval in which theorem guarantee unique solution.
2.  $( - 4, - 1),( - 1,1)$ and $(1,\infty )$  are interval in which theorem guarantee unique solution.

Step 1: Determine the interval for which theorem guarantee unique solution .

 $y^4-\ln x{y}^{\text{'}\text{'}}+x{y}^{\text{'}}+2y=\cos 3x$

Here 1,  $\ln x,x,2$ and  $\cos 3x$ have to be continuous.

Hence,

$(0,\infty )$  is interval in which theorem guarantee unique solution.

Step 2: Determine the interval for which theorem guarantee unique solution

$$\begin{gathered} \left(x^2-1\right)y^{\text{'}\text{'}\text{'}}+\sin x{y}^{\text{'}\text{'}}+\sqrt{x+4}{y}^{\text{'}}+e^{x}y=x^2+3 \\ {y}^{\text{'}\text{'}\text{'}}+\frac{\sin x}{\left(x^2-1\right)}{y}^{\text{'}\text{'}}+\frac{\sqrt{x+4}}{\left(x^2-1\right)}{y}^{\text{'}}+\frac{e^x}{\left(x^2-1\right)}y=\frac{x^2+3}{\left(x^2-1\right)} \end{gathered}$$

All quotient functions have to be continuous.

 $$\begin{gathered} x^2-1\ne 0\text{ and }\sqrt{x+4}>0 \\ x\in (-4,-1)\cup (-1,1)\cup (1,\infty ) \end{gathered}$$

Hence,

 $( - 4, - 1),( - 1,1)$ and  $(1,\infty )$ are interval in which theorem guarantee unique solution.