Question

Equations with the Dependent Variable Missing. For a scond order differential equation of the form \(y^{\prime \prime}=f\left(t, y^{\prime}\right),\) the substitution \(v=y^{\prime}, v^{\prime}=y^{\prime \prime}\) leads to a first order equation of the form \(v^{\prime}=f(t, v)\). If this equation can be solved for \(v\), then \(y\) can be obtained by integrating \(d y / d t=v .\) Note that one arbitrary constant is obtained in solving the first order equation for \(v,\) and a second is introduced in the integration for \(y\). In each of Problems 28 through 33 use this substitution to solve the given equation. $$ t^{2} y^{\prime \prime}+2 t y^{\prime}-1=0, \quad t>0 $$

Short answer

Answer: The general solution of the given second-order differential equation is \(y(t) = \ln{t} - \frac{1}{3}t^3 + C_1t + C_2\), where \(C_1\) and \(C_2\) are constants of integration.

Step-by-step solution

Perform the substitution

Substitute \(v=y'\) and \(v'=y''\) into the given equation. $$ t^{2}v'+2tv-1=0 $$

Solve for \(v'\)

Isolate \(v'\) in the equation: $$ v' = \frac{1-2tv}{t^2} $$

Integrate the equation to find \(v\)

Integrate the equation for \(v'\) with respect to \(t\): $$ v(t) = \int \frac{1-2tv}{t^2} dt $$ Using the power rule for integration, we find: $$ v(t) = \int \frac{1}{t^2} dt - 2\int t dt = -\frac{1}{t} - t^2 + C_1 $$ where \(C_1\) is the constant of integration.

Substitute back for \(y'\)

Now that we have found \(v(t)\), substitute back \(y'\) for \(v\). Recall that \(v=y'\): $$ y' = -\frac{1}{t} - t^2 + C_1 $$

Integrate to find \(y\)

Integrate the equation for \(y'\) with respect to \(t\) to find \(y(t)\): $$ y(t) = \int (-\frac{1}{t} - t^2 + C_1) dt $$ Using the power rule and factoring out the constant, we find: $$ y(t) = \ln{t} - \frac{1}{3}t^3 + C_1t + C_2 $$ where \(C_2\) is another constant of integration. So, the solution to the given second-order differential equation is: $$ y(t) = \ln{t} - \frac{1}{3}t^3 + C_1t + C_2 $$

Key concepts

Differential Equations

At the heart of many scientific and engineering problems lie differential equations. These equations express relationships between functions and their derivatives, describing how a quantity changes with respect to another.

In particular, second-order differential equations involve the second derivative of a function, and have the general form \( y'' = f(t, y') \). Such equations are pivotal in modeling scenarios like harmonic oscillators or electrical circuits. The goal is to find a function \( y(t) \) that satisfies the differential equation.

Understanding how the rate of change itself changes allows us to predict complex systems' behavior over time. Given the complexity involved, advanced methods like substitution are often necessary to find solutions.

Substitution Method

The substitution method is a powerful technique used to simplify solving differential equations, particularly when dealing with higher orders. The essence of substitution is to make the problem more approachable by introducing new variables.

For second-order equations, a common tactic is substituting \( v = y' \), transforming the equation into something more manageable. In our exercise, this reduces the original second-order equation to a first-order differential equation \( v' = \frac{1 - 2tv}{t^2} \).

This step crucially simplifies solving by focusing on \( v \), the first derivative of \( y \). Once \( v \) is determined, further integration can reveal \( y \), completing the solution process. The substitution method is not only about simplifying the calculation but also deepening understanding of the structure of differential equations.

Integration

Integration is a fundamental mathematical operation, essentially the reverse of differentiation. When solving differential equations, integration allows us to move from a derivative back to the original function.

When integrating, we often apply methods like the power rule to solve for the antiderivative. In our exercise, after rearranging to find \( v(t) \), we use integration to deduce \( v(t) = -\frac{1}{t} - t^2 + C_1 \).

This integration step is repeated to find \( y(t) \) from \( y' = v \), resulting in \( y(t) = \ln{t} - \frac{1}{3}t^3 + C_1t + C_2 \). Successful integration hinges on recognizing patterns and applying rules accurately.

Constants of Integration

When integrating, especially in the context of differential equations, constants of integration arise naturally. These constants account for the arbitrary shift or scaling that is possible with indefinite integrals.

When solving for functions, like \( v(t) \) or \( y(t) \), each integration introduces its constant. In our problem, \( C_1 \) and \( C_2 \) are these constants, reflecting how the solution can shift based on boundary or initial conditions.

These constants are not arbitrary in practice. They are determined through additional information provided in the problem, such as initial values or specific conditions. They personalize the general solution to fit specific scenarios, making each solution uniquely applicable.