Problem 7 - Advanced Engineering Mathematics

Question

Letting \(y=u x\) we have $$\begin{aligned} (u x-x) d x-(u x+x)(u d x+x d u) &=0 \\ \left(u^{2}+1\right) d x+x(u+1) d u &=0 \\ \frac{d x}{x}+\frac{u+1}{u^{2}+1} d u &=0 \\ \ln |x|+\frac{1}{2} \ln \left(u^{2}+1\right)+\tan ^{-1} u &=c \\ \ln x^{2}\left(\frac{y^{2}}{x^{2}}+1\right)+2 \tan ^{-1} \frac{y}{x} &=c_{1} \\\ \ln \left(x^{2}+y^{2}\right)+2 \tan ^{-1} \frac{y}{x} &=c_{1}. \end{aligned}$$

Step-by-Step Solution

Answer

The final expression in terms of \(x\) and \(y\) is \(\ln (x^2+y^2) + 2 \tan^{-1}(\frac{y}{x}) = c_1\).

Step 1: Substitute y with ux

Start by substituting the variable \(y\) with \(ux\), where \(u = \frac{y}{x}\). Rewrite the expression in terms of \(x\) and \(u\).

Step 2: Expand and Simplify the Expression

Expand the original expression \((ux-x)dx - (ux+x)(udx+xdu) = 0\). This leads to \(-x(udx + xdu) + ux^2du = 0\). Simplify by grouping similar terms together.

Step 3: Rearrange the Differential Equation

Rearrange the expression into a form that allows separation of variables. After simplification, it should become: \((u^2 + 1)dx + x(u+1)du = 0\).

Step 4: Separate Variables

Separate variables by dividing each term by \(x(u^2 + 1)\), resulting in the integral form: \(\frac{dx}{x} + \frac{u+1}{u^2+1} du = 0\).

Step 5: Integrate Both Sides

Integrate both sides separately:- The left side integrates to \(\ln|x|\).- The right side integrates to \(\frac{1}{2} \ln(u^2 + 1) + \tan^{-1}u\).

Step 6: Combine Integrals and Solve for Constant

Combine the results of Step 5 to form:\(\ln|x| + \frac{1}{2} \ln(u^2 + 1) + \tan^{-1}u = c\) where \(c\) is a constant of integration.

Step 7: Substitute back to get solution in x and y

Substitute \(u = \frac{y}{x}\) back into the equation from Step 6 to get the expression: \(\ln x^2 (\frac{y^2}{x^2} + 1) + 2\tan^{-1}(\frac{y}{x}) = c_1\).

Step 8: Simplify the Final Expression

Simplify this equation to form:\(\ln (x^2 + y^2) + 2 \tan^{-1}(\frac{y}{x}) = c_1\).

Key Concepts

Separation of Variables

When solving differential equations, one common technique is the separation of variables. This method is particularly useful for solving ordinary differential equations that can be rewritten such that all terms involving one variable and its derivative appear on one side of the equation, while those involving the other variable are on the opposite side. This rearrangement allows each side of the equation to be integrated independently.

In the provided exercise, after substituting and simplifying, the expression is rewritten as \[ (u^2 + 1)dx + x(u + 1)du = 0. \] By dividing all terms by \(x(u^2 + 1)\), the equation can be expressed as \[ \frac{dx}{x} + \frac{u + 1}{u^2 + 1} du = 0, \] enabling separation of variables. Each side can then be integrated separately, effectively breaking down the problem into simpler integrals. This makes the process more manageable and often allows for a solution to be found in terms of elementary functions.

Separation of variables is not always applicable, but when it is, it breaks down complex differential equations into easier-to-solve parts. Ultimately, this technique simplifies the process of finding solutions that meet both the initial conditions and the characteristics dictated by the differential equation itself.

Integration Techniques

To solve the integrals obtained from the separation of variables, integration techniques are essential. The exercise involves two main integrals after separating variables: \[ \int \frac{dx}{x} \] and \[ \int \frac{u + 1}{u^2 + 1} du. \]

The first integral, \(\int \frac{dx}{x},\) can be solved using the basic natural logarithm rule, yielding \(\ln |x|.\)

For the second integral, \(\int \frac{u + 1}{u^2 + 1} du,\) it's necessary to break it down into simpler parts. This can be achieved by separating it into two distinct components: \[ \int \frac{u}{u^2 + 1} du + \int \frac{1}{u^2 + 1} du. \] The first part, \(\int \frac{u}{u^2 + 1} du,\) is solved with a straightforward substitution method, whereas the second part, \(\int \frac{1}{u^2 + 1} du,\) results in the arctangent function, because the derivative of \(\tan^{-1}(u)\) is \(\frac{1}{u^2 + 1}.\) Together, these integrations lead to \[ \frac{1}{2} \ln(u^2 + 1) + \tan^{-1} u. \]

Employing various integration techniques effectively allows us to find solutions to different parts of a differential equation, ensuring each component is solved efficiently and accurately.

Ordinary Differential Equations

Ordinary differential equations (ODEs) are equations that involve functions of only one independent variable and their derivatives. ODEs are extensively used to model various natural and engineering problems, allowing us to describe the behaviour of physical systems.

In the exercise provided, we are dealing with an ordinary differential equation where the solution is sought in terms of the variable \(x\) and its relationship with \(y.\) The solution process involves expressing the differential equation using a technique like separation of variables, integrating it, and then converting the solution back in terms of the original variables.

The solution process also involves substituting an expression \(y = ux,\) which helps in simplifying the equation into a manageable form where functions of \(x\) and \(y\) can be separated and integrated independently. The successful application of these techniques results in an expression where, finally, the equation represents the combined effects of \(x\) and \(y.\) \[ \ln(x^2 + y^2) + 2 \tan^{-1} \left(\frac{y}{x}\right) = c_1. \]

Understanding ordinary differential equations and their solutions is key in mathematical modelling, providing insight into patterns and predicting behavior in real-world systems.