Question

Consider the initial-value problem $\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{x}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$. Determine which of the two curves shown in Figure 1.2.11 is the only plausible solution curve. Explain your reasoning.

Figure 1.2.11 Graphs for Problem 47

Short answer

• The y value will be $y=\frac{2x-1}{4}+\frac{3}{4e^{2x}}$
• The red curve is the only plausible solution curve.

Step-by-step solution

Definition of initial-value problem

The unknown function $y\left(x\right)$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.

Formula used for the solution

First order linear DE has the form of

$y\text{'}\left(x\right)+p\left(x\right)y=q\left(x\right)$

Determine the curves and find the y value

Here, the first order linear DE:

$y\text{'}\left(x\right)+p\left(x\right)y=q\left(x\right)$

By rewriting the given equation $y\text{'}=x-2y$using above formula

$y\text{'}+2y=x$

Where$p\left(x\right)=2,q\left(x\right)=x$

Now using the integrating factor

$I=e^{\int \left(p\left(x\right)dx\right)}$

Integrating\[\]

$e^{\int 2dx}=e^{2x}$

Now let’s multiply the DE with I

$e^{2x}y\text{'}+2y{e}^{2x}=e^{2x}x$

By solving the differential equation, we get

$y=\frac{2x-1}{4}+\frac{C}{e^{2x}}$

Step 4:Find the value of C

The graph passes through $\left(0,\frac{1}{2}\right)$

Substituting this in the y equation we get the value C

$$\begin{gathered} y=\frac{2x-1}{4}+\frac{C}{e^{2x}} \\ \frac{1}{2}=\frac{2\left(0\right)-1}{4}+\frac{C}{e^{2\left(0\right)}} \\ C=\frac{3}{4} \end{gathered}$$

So now,we get

$y=\frac{2x-1}{4}+\frac{3}{4e^{2x}}$

Hence the solution is$y=\frac{2x-1}{4}+\frac{3}{4e^{2x}}$

Graph

Hence, the Red curve is the plausible solution curve.