Question

Determine the plausible value of $\mathbf{ }{\mathit{x}}_{\mathbf{0}}$for which the graph of the solution of the initial value problem$\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{3}\mathit{x}\mathbf{-}\mathbf{6}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}{\mathit{x}}_{\mathbf{0}}\mathbf{\right)}\mathbf{=}\mathbf{0}$is tangent to x-axis at$\mathbf{(}{\mathit{x}}_{\mathbf{0}}\mathbf{,}\mathbf{0}\mathbf{)}$. Explain your reasoning.

Short answer

$x_0=2$

Step-by-step solution

Definition of initial-value problem

The unknown function y(x) and its derivatives at a number $x_0$. On some interval I containing $x_0$the problem of solving an nth-order differential equation subject to n side conditions specified at:

Solve: $\frac{d^{n}y}{d{x}^n}=f(x,y,y\text{'},...,y^{(n-1)})$

Subject to: $y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,y^{(n-1)}\left(x_0\right)=y_{(n-1)}.$

Where $y_0,y_1,...,y_{n-1}$are arbitrary constants, is called n-th order Initial Value Problem (IVP). The values of y(x) and its first n-1 derivatives at $x_0$$y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,y^{(n-1)}\left(x_0\right)=y_{(n-1)}.$ are called Initial Conditions.

Step2: Solving the equation

At the point $x=x_0$,if $y\left(x_0\right)$is tangent to x-axis

$y\text{'}\left(x_0\right)$is the slope of the tangent to x-axis

We have to satisfy that$y\text{'}\left(x_0\right)=0$

The slope of the x-axis is zero

Now substitute $y\text{'}\left(x_0\right)=0$in$y\text{'}+2y=3x-6$

And also $(x_0,0)$to$y_0=0$

We get,

$$\begin{gathered} y\text{'}\left(x_0\right)+2y_0=3x_0-6 \\ 0+2\left(0\right)=3x_0-6 \\ 0=3x_0-6 \\ 3x_0=6 \\ {x}_0=2 \end{gathered}$$

A plausible value of $x_0$for which the graph of the solution of the initial value problem $y\text{'}+2y=3x-6,y\left(x_0\right)=0$is tangent to x-axis at $(x_0,0)$is 2

Hence the solution is$x_0=2$