Problem 49 - Advanced Engineering Mathematics

Question

Use the substitution \(t=-x\) to solve the given initial-value problem on the interval \((-\infty, 0)\). $$ 4 x^{2} y^{\prime \prime}+y=0, y(-1)=2, y^{\prime}(-1)=4 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = 2 (-x)^{1/2} - 5 (-x)^{1/2} \ln(-x) \).

Step 1: Substitution

We start by using the substitution \( t = -x \). Under this substitution, we have \( x = -t \). Therefore, the derivatives transform as follows: \( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = -\frac{dy}{dt} \) \( \frac{d^2y}{dx^2} = \frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{d}{dt}\left( -\frac{dy}{dt} \right) \cdot \frac{dt}{dx} = \frac{d^2y}{dt^2} \)},

Step 2: Rewrite the Differential Equation

Substitute these derivative transformations into the differential equation:\[ 4(-t)^2 \frac{d^2y}{dt^2} + y = 0 \]This simplifies to\[ 4t^2 \frac{d^2y}{dt^2} + y = 0 \]

Step 3: Solve the Transformed Equation

Notice that this is a Cauchy-Euler type equation. By assuming a solution of the form \( y = t^m \), find the characteristic equation:\[ 4t^2 m(m-1) t^{m-2} + t^m = 0 \]This simplifies to \( 4m(m-1) + 1 = 0 \), yielding \( 4m^2 - 4m + 1 = 0 \). Factor or use the quadratic formula to get \( m = \frac{1}{2} \) (double root).

Step 4: General Solution for Double Root

Since we have a double root \( m = \frac{1}{2} \), the general solution for the differential equation is:\[ y(t) = C_1 t^{1/2} + C_2 t^{1/2} \ln(t) \]

Step 5: Revert to Original Variables

Convert back to the original variable \( x = -t \) (hence \( t = -x \)) to get:\[ y(x) = C_1 (-x)^{1/2} + C_2 (-x)^{1/2} \ln(-x) \]

Step 6: Apply Initial Conditions

We have initial conditions at \( x = -1 \): \( y(-1) = 2 \) and \( y'(-1) = 4 \).First condition gives \[ 2 = C_1 (1)^{1/2} + C_2 (1)^{1/2} \cdot \ln(1) = C_1 \]Thus, \( C_1 = 2 \).For the second derivative: \[ y'(t) = \frac{d}{dt}\left(C_1 t^{1/2} + C_2 t^{1/2} \ln(t)\right) \] \( y'(t) = C_1 \cdot \frac{1}{2} t^{-1/2} + C_2 \left( \frac{1}{2} t^{-1/2} \ln(t) + t^{-1/2} \right) \)Therefore at \( t = 1 \),\[ 4 = -\left( C_1 \cdot \frac{1}{2} + C_2 \cdot \frac{1}{2} \ln(1) + C_2 \right) = - 1 - C_2 \]Solving gives \( C_2 = -5 \).

Step 7: Final Solution

Combining the values of \( C_1 \) and \( C_2 \), the specific solution becomes:\[ y(x) = 2 (-x)^{1/2} - 5 (-x)^{1/2} \ln(-x) \]

Key Concepts

Initial-Value Problem

An initial-value problem involves finding a solution to a differential equation that satisfies certain specified values, known as initial conditions. In this case, the problem is to solve a second-order differential equation with given initial values at a specific point.
These initial conditions often help determine the constants in the general solution of the differential equation. For the problem given here, two initial conditions are specified:

• \( y(-1) = 2 \)
• \( y'(-1) = 4 \)

These conditions ensure that the solution not only satisfies the differential equation but also fits the physical or geometrical requirements of the problem at the given points.
Without these conditions, we can only derive a general solution incorporating arbitrary constants.

Change of Variables

The technique of change of variables is a powerful tool in calculus, especially for simplifying and solving differential equations. In this context, it involves substituting a new variable to transform the differential equation into a more tractable form.
For the original equation, the substitution is made with \( t = -x \). This substitution helps to reframe the problem across an easier domain for analysis.
Changing variables modifies both the derivatives and the independent variables in the equation. This new form often simplifies the relationships in the equation and makes it easier to solve. After solving in the new variable, one must convert back to the original for a complete solution.

Characteristic Equation

The characteristic equation emerges from assuming a solution to a differential equation of a specific form, often involving exponential or power functions. For Cauchy-Euler equations, a common tactic is to assume solutions of the form \( y = t^m \).
Substituting this form into the differential equation allows for deriving a characteristic equation, which is essentially an algebraic equation. Solving it gives the values of \( m \) that satisfy the original differential equation.
For this problem, substituting into the differential equation led to \[ 4m(m-1) + 1 = 0 \]which simplifies to \[ 4m^2 - 4m + 1 = 0 \].
Solving this characteristic equation provides the foundation of the general solution of the differential equation.

Double Root Solution

When the characteristic equation yields a double root, it indicates a special type of solution for the differential equation. Double roots mean both solutions are the same value and must be treated uniquely to form a complete general solution.
For a double root \( m = \frac{1}{2} \), the general solution takes on the form of a linear combination of the standard solution and a modified version with a logarithmic term. Hence, in this context, the solution becomes:
\[ y(t) = C_1 t^{1/2} + C_2 t^{1/2} \ln(t) \]
This adaptation ensures that the general solution covers the case of repeated solutions in the characteristic equation, allowing for a wider range of behavior in the solution set.