Problem 4 - Advanced Engineering Mathematics

Question

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

Step-by-Step Solution

Answer

The solution is \( y = -\ln|\cos(x + C)| + C' \).

Step 1: Substitute the Given Values

Start by substituting \( u = y' \). This makes the differential equation \( y'' = 1 + (y')^2 \) become \( \frac{du}{dx} = 1 + u^2 \).

Step 2: Separate the Variables

Rearrange the equation from Step 1 to isolate \( u \) and \( x \) by moving all terms involving \( u \) to one side: \( \frac{du}{1 + u^2} = dx \).

Step 3: Integrate Both Sides

Integrate both sides: \( \int \frac{du}{1 + u^2} = \int dx \). The left side results in \( \tan^{-1}(u) \) and the right side results in \( x + C \), where \( C \) is the constant of integration. This gives: \( \tan^{-1}(u) = x + C \).

Step 4: Solve for \( u \)

Solve \( \tan^{-1}(u) = x + C \) to find \( u \). By taking the tangent of both sides, we get: \( u = \tan(x + C) \).

Step 5: Substitute Back for \( y' \)

Replace \( u \) with \( y' \) to find: \( y' = \tan(x + C) \).

Step 6: Integrate to Find \( y \)

Integrate \( y' = \tan(x + C) \) to find \( y \). Use the identity \( \int \tan(x) \, dx = -\ln|\cos(x)| + C' \): thus, \( y = -\ln|\cos(x + C)| + C' \). Note there could be an additional constant \( C' \) from this integration.

Key Concepts

Substitution Method

The substitution method is a way to simplify complex differential equations by changing variables. In our exercise, we used the substitution \( u = y' \). This means we let \( u \) represent the derivative of \( y \), denoted by \( y' \).
This substitution turns a second-order differential equation into a first-order equation in terms of \( u \).
This makes it easier to solve.

The steps involve:

• Identifying an appropriate substitution that simplifies the equation.
• Rewriting the original differential equation in terms of the new variable.

By doing this, we transformed \( y'' = 1 + (y')^2 \) into \( \frac{du}{dx} = 1 + u^2 \).
This is simpler since it involves only \( u \) and \( x \).
Our goal is to solve for \( u \), then substitute back to find \( y'\) or \( y \).

Integration

Integration is a fundamental concept used to solve differential equations. It helps in finding a function from its derivative.
In our exercise, once we had \( \frac{du}{1 + u^2} = dx \), we integrated both sides.
This gives us the integral \( \int \frac{du}{1 + u^2} = \int dx \).

The integration of \( \int \frac{du}{1 + u^2} \) is a standard integral, known to result in the inverse tangent function:

• \( \int \frac{du}{1 + u^2} = \tan^{-1}(u) \)

This helps us simplify the equation to \( \tan^{-1}(u) = x + C \).
On the right, \( \int dx \) is simply \( x + C \), where \( C \) is an integration constant that arises because the integral of a constant is not fixed.
Thus, integration allowed us to find a relationship between \( u \) and \( x \).
This step is crucial for transforming derivatives back into functions.

Separation of Variables

Separation of variables is a technique used to rearrange a differential equation so that each variable is on a different side of the equation.
In our problem, after substituting \( u = y' \), we wanted to isolate \( u \) and \( x \).
We rearranged the simplified equation \( \frac{du}{dx} = 1 + u^2 \) to become \( \frac{du}{1 + u^2} = dx \).

The process involves:

• Moving terms involving \( u \) (like \( du \)) to one side.
• Keeping terms involving \( x \) (like \( dx \)) on the other side.

This allows us to integrate each side separately.
By separating the variables, we ensured that integration could be performed independently, helping us to solve the differential easily.
This method is powerful for first-order differential equations, where it’s possible to express the solution explicitly in terms of the variables themselves.

Second-Order Differential Equations

Second-order differential equations involve derivatives of a function up to the second degree, denoted as \( y'' \). These equations can describe systems with accelerations, such as motion under constant force.
In our exercise, the given second-order differential equation was \( y'' = 1 + (y')^2 \).

Second-order equations often need more complex solving techniques compared to first-order ones:

• They can require substitution to reduce their order, making them resemble first-order equations.
• The solutions can often describe curves or physical phenomena like oscillations.

By substituting \( u = y' \), we reduced the complexity by transforming the second-order differential equation into a first-order one in terms of \( u \).
This step is vital as it allows us to apply simpler methods like separation of variables and basic integration to solve the equation.