Question

Find the form of the solutions of the equation $$ \frac{d y}{d x} \frac{d^{3} y}{d x^{3}}-2\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{2}=0 $$ which have \(y(0)=\infty\) (You will need the result \(\int^{z} \operatorname{cosech} u d u=-\ln (\operatorname{cosech} z+\operatorname{coth} z)\).)

Short answer

The solution form is \( y(x) = -\ln (\operatorname{cosech} (k x + \phi) + \operatorname{coth} (k x + \phi)) \).

Step-by-step solution

Identify the given differential equation

Given: \[ \frac{d y}{d x} \frac{d^{3} y}{d x^{3}}-2\left(\frac{d^{2} y}{d x^{2}}\right)^{2}+\left(\frac{d y}{d x}\right)^{2}=0 \]

Introduce substitution

Let \( v = \frac{d y}{d x} \). Then, \( v' = \frac{d^{2} y}{d x^{2}} \) and \( v'' = \frac{d^{3} y}{d x^{3}} \). Substitute into the equation.

Transform the equation

The equation becomes: \[ v v'' - 2(v')^2 + v^2 = 0 \]

Simplify the equation

Rewrite it as: \[ v v'' - 2(v')^2 + v^2 = 0 \implies v'' - 2 \frac{(v')^2}{v} + v = 0 \]

Introduce new substitution

Let \( w = v' \). Thus, \( \frac{dw}{dx} = v'' \). The equation now reads: \[ v \frac{dw}{dx} - 2 \frac{w^2}{v} + v = 0 \]

Solve the differential equation

We get: \[ v \frac{dw}{dx} = 2 \frac{w^2}{v} - v \]Dividing through by \( v \):\[ \frac{dw}{dx} = 2 \frac{w^2}{v^2} - 1 \]

Separate variables and integrate

Separating variables, we have: \[ \int \frac{v^2}{2w^2- v^2} \, dw = \int dx \]Using the result from the prompt:

Integrate the left-hand side

We use the integral formula: \[ \int_{z} \operatorname{cosech} \, u \, du = -\ln \left( \operatorname{cosech} \, z + \operatorname{coth} \, z \right) \]

Solve for w in terms of v

After integration, we find, \( w = c \operatorname{coth} \, u \), where \( c \) is the integration constant.

Solve for the original variables

Re-substitute to express in terms of \( y \) and \( x \): \( \frac{d y}{d x} = c \operatorname{coth} \, (k x + \phi) \)

Integrate again to find y

Integrate with respect to \( x \): \[ y(x) = \int c \operatorname{coth} \, (k x + \phi) \, dx \]\Using provided integral result.

Apply initial conditions

Given that \( y(0) = \infty \), we can solve for constants. Usually, this leads to expressing \( \phi \) for proper description.

Key concepts

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving a function and its derivatives. For example, the given differential equation \[ \frac{d y}{d x} \frac{d^{3} y}{d x^{3}}-2\bigg(\frac{d^{2} y}{d x^{2}}\bigg)^{2}+\bigg(\frac{d y}{d x}\bigg)^{2}=0 \] is an ODE because it involves the derivatives of the function \( y(x) \).

ODEs are crucial in modeling physical, biological, and chemical processes. Understanding how to solve these allows us to predict behavior under given conditions. Solutions to ODEs can be found using various methods, such as separation of variables, integrating factors, or substitution. Usually, solving an ODE involves finding the general solution and then applying initial conditions to find a particular solution.

Substitution Method

The substitution method simplifies ODEs by introducing new variables. For our problem, we let \( v = \frac{d y}{d x} \).

Then, differentiating \( v \) with respect to \( x \) gives \( v' = \frac{d^2 y}{d x^2} \) and differentiating once more gives \( v'' = \frac{d^3 y}{d x^3} \). We substitute these into the original equation to convert it into a more manageable form: \[ v v'' - 2(v')^2 + v^2 = 0 \]

The equation further simplifies when we introduce another substitution to handle the second derivative of \( y \). This iterative substitution and simplification process makes the equation easier to solve.

Initial Conditions

Initial conditions provide specific values at the start point of the problem. These are essential for finding particular solutions to an ODE. For our problem, the initial condition given is \( y(0) = \infty \).

Initial conditions help determine the constants of integration after obtaining the general solution. This allows us to uniquely identify the solution that applies to our scenario. Without applying these conditions, we could only solve for a general family of solutions.

Integration Techniques

Integration techniques are methods used to solve integrals that arise during the process of solving differential equations. In our problem, after transforming and simplifying the equation, we separate the variables and integrate both sides: \[ \frac{dw}{dx} = \frac{2 \frac{w^2}{v^2} - 1}{v} \]

This requires performing both direct integration and using known integral results. For example, in our case, we use the integral for the hyperbolic cosecant function: \[ \int \operatorname{cosech} \ u \ du = -\ln \left( \operatorname{cosech} \ z + \operatorname{coth} \ z \right) \]

We apply this integral result to express our solution in terms of the original variables. Integration techniques like substitution, integration by parts, and partial fractions are frequently used in solving ODEs. Each technique is chosen based on the form of the equation at hand.