Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Consider the differential equation  $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{x}\mathbf{+}\mathit{s}\mathit{i}\mathit{n}\mathbf{ }\mathit{y}$

⦁    A solution curve passes through the point $\left(1,\frac{\pi }{2}\right)$ . What is its slope at this point?

⦁    Argue that every solution curve is increasing for $\mathit{x}\mathbf{>}\mathbf{1}$ .

⦁    Show that the second derivative of every solution satisfies $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x} \mathbf{cos} \mathbf{y}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{sin} \mathbf{2}\mathbf{y}\mathbf{.}$ 

⦁    A solution curve passes through (0,0). Prove that this curve has a relative minimum at (0,0).

Step-by-Step Solution

Answer

⦁    The slope at the point 2.

⦁    Yes, every solution curve is increasing for x > 1.

⦁    The second derivative of every solution satisfies the given equation.

⦁    Yes, the curve has a minimum at (0,0)

1(a): Find the slope of solution curve at 1 , π 2

Slope is given by  $\frac{\mathbf{dy}}{\mathbf{dx}}$

So, slope of the solution curve at  $\left(\mathbf{1}\mathbf{,}\frac{\pi }{\mathbf{2}}\right)$ is

$$\begin{gathered} \frac{\mathbf{dy}}{\mathbf{dx}}\left|\left(\mathbf{1}\mathbf{,}\frac{\pi }{\mathbf{2}}\right)\right.\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{sin}\left(\mathbf{1}\mathbf{,}\frac{\pi }{\mathbf{2}}\right) \\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{1} \\ \mathbf{=}\mathbf{2} \end{gathered}$$

Hence, the slope at the point is 2.

2(b): Compute dy dx for x > 1

Since,  $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{y}$ for all  $\mathbf{y}\in \mathrm{ℝ}$

Then, $\mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{y}\mathbf{>}\mathbf{1}$ , for all $\mathbf{x}\mathbf{>}\mathbf{1}$.

i.e., $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{>}\mathbf{1}\mathbf{>}\mathbf{0}$ , for all $\mathbf{x}\mathbf{>}\mathbf{1}$, $\mathbf{y}\in \mathrm{ℝ}$.

Hence, from first derivative test, every solution curve is increasing for $\mathbf{x}\mathbf{>}\mathbf{1}$ .

3(c): Determine the second derivative.

Here 

 $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{y}$

Differentiate both sides with respect to x

$$\begin{gathered} \frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{cos} \mathbf{y}\left(\frac{\mathbf{dy}}{\mathbf{dx}}\right) \\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{cos} \mathbf{y}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{y}\mathbf{)} \\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x} \mathbf{cos} \mathbf{y}\mathbf{+}\mathbf{sin} \mathbf{y} \mathbf{cos} \mathbf{y} \\ \mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x} \mathbf{cos} \mathbf{y}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{sin} \mathbf{2}\mathbf{y} \end{gathered}$$

So, the second derivative of every solution satisfies the given equation.

4(d): Find second derivative at ( 0 , 0 )

Since,  $\mathbf{x}\mathbf{+}\mathbf{sin} \mathbf{y}\mathbf{=}\mathbf{0}$ at  $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$

we get,  $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{0}$ at $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$ 

thus, $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$  is a critical point.

Also, from part (c) , $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{x} \mathbf{cos} \mathbf{y}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{sin} \mathbf{2}\mathbf{y}$

 $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{>}\mathbf{0}$ at  $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$

From second derivative test,  $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$ is a point of relative minimum.

Therefore, the curve has a minimum at  $\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{0}\mathbf{)}$