Problem 6 - Advanced Engineering Mathematics

Question

In Problems, solve the given differential equation by using the substitution \(u=y^{\prime}\). $$ (y+1) y^{\prime \prime}=\left(y^{\prime}\right)^{2} $$

Step-by-Step Solution

Answer

The general solutions are: \( y = A e^{kx} - 1 \) and \( y = B e^{-kx} - 1 \).

Step 1: Substitution

Let us substitute \(u = y'\), which means \(u = \frac{dy}{dx}\). Then the second derivative \( y'' \) can be written in terms of \( u \) as a derivative with respect to \( y \). Thus, \( y'' = \frac{du}{dx} = \frac{du}{dy}\frac{dy}{dx} = u \frac{du}{dy} \).

Step 2: Rewrite the Equation

Substitute the expressions for \( y' \) and \( y'' \) into the differential equation, replacing \( y' = u \) and \( y'' = u \frac{du}{dy} \). The equation becomes \( (y+1)u \frac{du}{dy} = u^2 \).

Step 3: Simplify the Equation

Cancel \( u \) from both sides of the equation since \( u eq 0 \):\[(y+1) \frac{du}{dy} = u.\]

Step 4: Separate the Variables

Rearrange the equation to separate the variables \( u \) and \( y \):\[ \frac{du}{u} = \frac{dy}{y+1}.\]

Step 5: Integrate Both Sides

Integrate both sides of the equation:\[ \int \frac{du}{u} = \int \frac{dy}{y+1}.\]The result is:\[ \ln |u| = \ln |y+1| + C,\]where \( C \) is the integration constant.

Step 6: Solve for \( u \)

Exponentiate both sides to solve for \( u \):\[ |u| = e^C |y+1|.\]Let \( e^C = k \), where \( k > 0 \), then:\[ u = k(y+1) \text{ or } u = -k(y+1).\]

Step 7: Back-Substitute for \( u \)

Since \( u = y' \), we have:\[ y' = k(y+1) \text{ or } y' = -k(y+1).\]These are first-order separable differential equations to solve for \( y \).

Step 8: Solve the Differential Equation

Separate the equations and solve:For \( y' = k(y+1) \):\[ \frac{dy}{y+1} = k dx.\]Integrate both sides:\[ \ln |y+1| = kx + C_1.\]For \( y' = -k(y+1) \):\[ \frac{dy}{y+1} = -k dx.\]Integrate both sides:\[ \ln |y+1| = -kx + C_2.\]

Step 9: Solve for \( y \) in Both Cases

In both cases:1. \( y = A e^{kx} - 1 \) where \( A = e^{C_1} \).2. \( y = B e^{-kx} - 1 \) where \( B = e^{C_2} \).These represent the general solutions for the differential equation.

Key Concepts

Substitution Method

One powerful method to simplify and solve complex differential equations is **Substitution**. When dealing with derivatives, sometimes direct solving isn't feasible without first transforming the equation. This is where substitution comes in handy. In the exercise above, we are given a complicated second-order differential equation, \[(y+1) y''=(y')^2\] We make the substitution \( u = y' \),which simplifies the second derivative to \( y'' = u \frac{du}{dy} \). It's like renaming parts of the equation temporarily to make it easier to work with. This transformation reduces the complexity by one order, changing it into a first-order equation. Substitution turns a problem into a new form that is simpler to solve:

• Rewriting derivatives enables easier manipulation.
• Allows the application of integration and other solving techniques on simpler equations.

Once obtained, the substitution allows us to back-substitute the original variables, returning to the context of the problem and thus achieving the solution.

Separable Differential Equations

A differential equation is **Separable** if the variables can be rearranged so each appears on different sides of the equation. This separation is crucial as it allows us to integrate each side independently. In the simplified differential equation from our problem \[(y+1) \frac{du}{dy} = u\], we can separate variables \( \frac{du}{u} = \frac{dy}{y+1} \). This form indicates that the equation is ready for integration, one of the key advantages of converting it to a separable form.Here's what to look for:

• Identify terms involving only one variable and move them to one side of the equation.
• Ensure each side of the equality is expressed in terms of a single variable.

This technique leverages the fundamental idea of separating to pave the way for integration, which yields the solution to the differential equation.

Integration Techniques

**Integration** is the process used to find a function given its derivative, a crucial aspect when dealing with differential equations. After separating variables, the next step is to integrate both sides independently:\[ \int \frac{du}{u} = \int \frac{dy}{y+1} \].The integration results in: \[ \ln |u| = \ln |y+1| + C \], where \(C\) is the integration constant. When integrating:

• The indefinite integral allows for the inclusion of a constant \(C\), representing an infinite family of solutions.
• Integration transforms differential equations into algebraic equations, leading to the final solution.

Solving these integrals conveniently uncovers the behavior of functions described by the original differential equation. Finally, remember that after integrating and simplifying, exponentiate both sides if needed, as shown in our exercise, to clear logarithmic terms, arriving at the general solution.