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. $$ y^{\prime \prime}+t\left(y^{\prime}\right)^{2}=0 $$

Short answer

Question: Transform the given second-order differential equation into a first-order equation and solve for \(y\) given the equation \(y^{\prime \prime} + t\left(y^{\prime}\right)^{2} = 0\). Answer: The solution to the given equation is \(y = -\ln\left|\frac{1}{2}t^2 + C_1\right| + C_2\).

Step-by-step solution

Substitution

Replace \(y^{\prime}\) with \(v\) and \(y^{\prime \prime}\) with \(v^\prime\): $$v^{\prime} + t\,v^{2} = 0$$ Now we have the first-order differential equation in the form \(v^{\prime}=f(t, v)\).

Separate the variables and rewrite the equation

In order to solve the first-order differential equation, we need to separate the variables. Divide by \(v^2\) and multiply by \(-1\) to get: $$\frac{-v^{\prime}}{v^2} = t$$

Integrate both sides of the equation

Now that we have separated the variables, we can integrate both sides of the equation: $$\int \frac{-1}{v^2} dv = \int t\,dt$$ On the left-hand side, we get: $$-\int v^{-2} dv = -\left[\frac{1}{v}\right]$$ On the right-hand side, we get: $$\int t\, dt = \frac{1}{2}t^2 + C_1$$ So, we have: $$-\frac{1}{v} = \frac{1}{2}t^2 + C_1$$

Solve for \(v\)

Now, we will solve for \(v\) by taking the reciprocal of both sides and multiplying by -1: $$v = -\frac{1}{\frac{1}{2}t^2 + C_1}$$ Now, recall that \(v = \frac{dy}{dt}\).

Integrate to solve for \(y\)

Now that we have found \(v\), we can integrate to find \(y\). First, note that \(v = \frac{dy}{dt}\), so we have: $$\frac{dy}{dt} = -\frac{1}{\frac{1}{2}t^2 + C_1}$$ Integrate both sides: $$\int dy = -\int \frac{1}{\frac{1}{2}t^2 + C_1} dt$$ Let \(u = \frac{1}{2}t^2 + C_1\) and substitute. Then, \(du = tdt\), and we get: $$\int dy = -\int \frac{1}{u} du$$ Now we can integrate: $$y = -\ln|u| + C_2$$ Substitute back for \(u\) and obtain the final solution for \(y\): $$y = -\ln\left|\frac{1}{2}t^2 + C_1\right| + C_2$$ Thus, the solution to our initial second-order differential equation is: $$y = -\ln\left|\frac{1}{2}t^2 + C_1\right| + C_2$$

Key concepts

First-Order Differential Equations

A first-order differential equation involves derivatives of the first degree and is typically in the form \( y' = f(t, y) \). The main advantage of converting a second-order differential equation to a first-order one, as demonstrated in the exercise, is simplicity. It allows us to handle the problem with a more straightforward approach, focusing directly on two variables instead of three.
Often in these kinds of problems, the method involves replacing the higher-order derivatives. For our exercise, the substitution process simplified the equation to \( v' + tv^2 = 0 \). This manipulable form makes solving the differential equation more straightforward and consistent with the techniques available for first-order equations.

Variable Substitution

Variable substitution is a strategic method used to simplify complex differential equations. In the context of our problem, we substitute \( y' \) with \( v \) and \( y'' \) with \( v' \). This transforms the equation from a second-order form to a first-order one.
By applying this technique, we simplify the process into more manageable parts. After substitution, the equation \( v' + t v^2 = 0 \) becomes evident, paving the way for subsequent steps. This substitution is crucial as it reduces the cognitive load and allows the focus to shift towards solving a more straightforward mathematical problem. In turn, it enables the utilization of various solving techniques inherent to first-order differential equations.

Integration Methods

Integration is the process of finding the antiderivative of a function, which is integral to solving differential equations. After separating variables in our transformed first-order equation, we use integration on both sides to find a solution.
The expression \( \int \frac{-1}{v^2} dv = \int t\,dt \) shows the method of integrating each part. The left side simplifies to \( -\left[\frac{1}{v}\right] \), and the right side yields \( \frac{1}{2}t^2 + C_1 \). These integrations help in finding the relationship between the variables, crucial for eventually determining an explicit function for \( v \).
After determining \( v \), we further integrate \( \frac{dy}{dt} = v \) to obtain \( y \), which involves using integration techniques once more with the substitution methods explained.

Arbitrary Constants

Arbitrary constants are constants that arise during the integration of differential equations. For instance, when we solve \( \int t\, dt \), we add \( C_1 \), an arbitrary constant. Similarly, a second arbitrary constant, \( C_2 \), appears when integrating to solve for \( y \).
These constants are essential because they represent the general nature of the solution. Since integration is an indefinite process, these constants account for all possible curves that the solution might take. In practical problems, these constants can be interpreted or specified using initial conditions or boundary values.
In our equation, adding \( C_1 \) and \( C_2 \) showcases the presence of two constants due to double integration from the second-order equation, ensuring a comprehensive solution set that caters to various possibilities.