Problem 18 - Advanced Engineering Mathematics

Question

(a) Writing the equation in the form \((x-\sqrt{x^{2}+y^{2}}) d x+y d y=0\) we identify \(M=x-\sqrt{x^{2}+y^{2}}\) and \(N=y\) since \(M\) and \(N\) are both homogeneous functions of degree 1 we use the substitution \(y=u x .\) It follows that \\[ \begin{aligned} (x-\sqrt{x^{2}+u^{2} x^{2}}) d x+u x(u d x+x d u) &=0 \\ x\left[1-\sqrt{1+u^{2}}+u^{2}\right] d x+x^{2} u d u &=0 \\ -\frac{u d u}{1+u^{2}-\sqrt{1+u^{2}}} &=\frac{d x}{x} \\ \frac{u d u}{\sqrt{1+u^{2}}(1-\sqrt{1+u^{2}})} &=\frac{d x}{x} \end{aligned} \\] Letting \(w=1-\sqrt{1+u^{2}}\) we have \(d w=-u d u / \sqrt{1+u^{2}}\) so that \\[ \begin{aligned} -\ln |1-\sqrt{1+u^{2}}| &=\ln |x|+c \\ \frac{1}{1-\sqrt{1+u^{2}}} &=c_{1} x \\ 1-\sqrt{1+u^{2}} &=-\frac{c_{2}}{x} \\ 1+\frac{c_{2}}{x} &=\sqrt{1+\frac{y^{2}}{x^{2}}} \\ 1+\frac{2 c_{2}}{x}+\frac{c_{2}^{2}}{x^{2}} &=1+\frac{y^{2}}{x^{2}} \end{aligned} \\] Solving for \(y^{2}\) we have \\[ y^{2}=2 c_{2} x+c_{2}^{2}=4\left(\frac{c_{2}}{2}\right)\left(x+\frac{c_{2}}{2}\right) \\] which is a family of parabolas symmetric with respect to the \(x\) -axis with vertex at \(\left(-c_{2} / 2,0\right)\) and focus at the origin. (b) Let \(u=x^{2}+y^{2}\) so that \\[ \frac{d u}{d x}=2 x+2 y \frac{d y}{d x} \\] Then \\[ y \frac{d y}{d x}=\frac{1}{2} \frac{d u}{d x}-x \\] and the differential equation can be written in the form \\[ \frac{1}{2} \frac{d u}{d x}-x=-x+\sqrt{u} \text { or } \frac{1}{2} \frac{d u}{d x}=\sqrt{u} \\] Separating variables and integrating gives $$\begin{aligned} \frac{d u}{2 \sqrt{u}} &=d x \\ \sqrt{u} &=x+c \\ u &=x^{2}+2 c x+c^{2} \\ x^{2}+y^{2} &=x^{2}+2 c x+c^{2} \\ y^{2} &=2 c x+c^{2} \end{aligned}$$

Step-by-Step Solution

Answer

Part (a): Family of parabolas symmetric about x-axis with vertex at \((-c_2/2, 0)\). Part (b): Same parabola equation \(y^2 = 2cx + c^2\).

Step 1: Identify Homogeneous Functions

First, we identify the homogeneous functions from the given differential equation. We have \(M = x - \sqrt{x^2 + y^2}\) and \(N = y\). Both \(M\) and \(N\) are homogeneous functions of degree 1, indicating we can use the substitution \(y = ux\).

Step 2: Substitution and Simplify

Using the substitution \(y = ux\), the differential equation transforms to \((x - \sqrt{x^2 + u^2x^2})dx + ux(udx + xdu) = 0\). Simplifying gives \(x[1 - \sqrt{1 + u^2} + u^2]dx + x^2 u du = 0\).

Step 3: Separate Variables

Rearranging terms leads to \(-\frac{u du}{1 + u^2 - \sqrt{1 + u^2}} = \frac{dx}{x}\). Further simplifying gives \(\frac{u du}{\sqrt{1 + u^2}(1 - \sqrt{1 + u^2})} = \frac{dx}{x}\).

Step 4: Integration with Substitution

Introduce a substitution \(w = 1 - \sqrt{1 + u^2}\), leading to \(dw = -\frac{u du}{\sqrt{1 + u^2}}\). The equation becomes \(-\ln|1 - \sqrt{1 + u^2}| = \ln|x| + c\).

Step 5: Express Family of Curves

After further manipulation, we arrive at \(1 - \sqrt{1 + u^2} = -\frac{c_2}{x}\). Solving this gives \(y^2 = 2c_2x + c_2^2\), describing a family of parabolas symmetric about the x-axis.

Step 6: Part (b) - Define New Variable

For part (b), define \(u = x^2 + y^2\), giving \(\frac{du}{dx} = 2x + 2y\frac{dy}{dx}\) and subsequently, \(\frac{dy}{dx} = \frac{1}{2y}(\frac{du}{dx} - 2x)\).

Step 7: Rewrite Differential Equation

Substitute to rewrite the differential equation: \(y \frac{dy}{dx} = \frac{1}{2}\frac{du}{dx} - x\). Simplifying gives \(\frac{1}{2}\frac{du}{dx} = \sqrt{u}\).

Step 8: Separate and Integrate (Part b)

Separate variables: \(\frac{du}{2\sqrt{u}} = dx\). Integrate this to obtain \(\sqrt{u} = x + c\).

Step 9: Find Parabola Equation (Part b)

Square both sides and substitute back: \(u = (x + c)^2\). Therefore, \(x^2 + y^2 = x^2 + 2cx + c^2\) leading to \(y^2 = 2cx + c^2\).

Key Concepts

Homogeneous Functions

Homogeneous functions are important in solving certain types of differential equations. In the given example, we identify the functions \(M = x - \sqrt{x^2 + y^2}\) and \(N = y\). Both of these functions are homogeneous functions of degree 1. This simply means that if we multiply all variables by a constant \(t\), the function also gets multiplied by \(t\). For instance, \(M(tx, ty) = tM(x,y)\). Recognizing this property allows us to apply specific techniques, like variable substitution, to simplify the differential equation. Homogeneous functions help to handle equations that would otherwise seem too complex to solve directly.

Variable Substitution

When dealing with differential equations, variable substitution is a powerful tool to simplify the problem. In this problem, we use the substitution \(y = ux\). Here, \(u\) is defined as the new variable that represents the ratio of \(y\) to \(x\). This substitution helps in reducing the complexity of the equation. After substitution, the differential equation transforms into a form that makes it easier to rearrange and simplify. By doing so, we can separate the variables or simplify the integration process. The substitution essentially provides a new perspective that often reveals simpler relationships.

Separation of Variables

Separation of variables is a method used in solving differential equations by separating the variables in such a way that all terms involving one variable are on one side of the equation and all terms involving the other variable are on the other side. In our case, after substitution, we obtain:

• \( -\frac{u \ du}{1 + u^2 - \sqrt{1 + u^2}} = \frac{dx}{x}\)
• This form allows us to integrate both sides independently.

The separation makes integration straightforward because we can address each side separately. This technique is particularly useful in ordinary differential equations like the one we're handling, enabling the equation to be solved much more efficiently.

Parabolic Equations

Parabolic equations often describe curves that have a specific geometric shape. In the exercise, the solution leads to identifying that the curves are parabolas. Through manipulation and simplification, we find that the equation \( y^2 = 2c_2x + c_2^2 \) describes a family of parabolas. These parabolas are symmetric with respect to the x-axis with their vertex located at \(\left(-\frac{c_2}{2}, 0\right)\) and focus at the origin. The fascinating outcome here is seeing how a seemingly complex differential equation can beautifully describe such simple geometric shapes. Parabolic equations are prevalent in physics and engineering, modeling paths like projectiles or antenna shapes.