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. $$ y^{\prime \prime}+y^{\prime}=e^{-t} $$

Short answer

To solve the given second-order differential equation, we first transformed it into a first-order differential equation by substituting v for y' and v' for y''. We then solved the first-order differential equation to find v and integrated v with respect to t to find y. The final solution is: $$ y(t) = -(t+C_1)e^{-t} - e^{-t} + C_2 $$ where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Substituting v and v' for y' and y'' respectively

Given the second order differential equation: $$ y^{\prime\prime} + y^{\prime} = e^{-t} $$ Let's make the substitution \(v = y^{\prime}\), and \(v^{\prime} = y^{\prime\prime}\). The given equation becomes: $$ v^{\prime} + v = e^{-t} $$ Now we have a first-order differential equation in terms of \(v\) and \(t\).

Solve the first-order differential equation for v

The first-order differential equation in terms of \(v\) and \(t\) is: $$ v^{\prime} + v = e^{-t} $$ This is a first-order linear differential equation. To solve it, we need to find an integrating factor. The integrating factor \(I(t)\) for this equation is given by: $$ I(t) = \exp\left(\int 1 \, dt \right) = e^{t} $$ Now, multiply the equation by the integrating factor: $$ e^t v^{\prime} + e^t v = e^{-t} e^t $$ $$ \frac{d}{dt}\left(e^t v\right) = 1 $$ Now, integrate both sides with respect to t: $$ \int\frac{d}{dt}\left( e^t v\right) dt = \int 1 dt $$ $$ e^t v = t + C_1 $$ Now, isolate \(v\): $$ v = e^{-t}(t + C_1) $$

Integrate v to find y

Recall that \(v = y^{\prime}\). So now we need to integrate \(v\) with respect to \(t\) to find \(y\): $$ y^{\prime} = e^{-t}(t + C_1) $$ $$ y(t) = \int e^{-t}(t + C_1) dt $$ To integrate this expression, we can use integration by parts, let: $$ u = t + C_1,\ dv = e^{-t} dt \\ du = dt,\ v = -e^{-t} $$ Applying integration by parts formula (\(\int u dv = uv - \int v du\)), we get $$ y(t) = -(t+C_1)e^{-t} + \int e^{-t} dt $$ $$ y(t) = -(t+C_1)e^{-t} - e^{-t} + C_2 $$ Therefore, the solution to the original second-order differential equation is: $$ y(t) = -(t+C_1)e^{-t} - e^{-t} + C_2 $$ where \(C_1\) and \(C_2\) are arbitrary constants.

Key concepts

First Order Differential Equations

First order differential equations are equations that involve first derivatives of a function. They have a general form: \(\frac{dy}{dt} = f(t, y)\). The essence lies in the dependency of the rate of change of a variable on the variable itself and possibly on time or another independent variable. These equations are quite common and can represent many real-life processes such as exponential growth or decay, Newton's Law of Cooling, and more.
A key approach to solve a first order differential equation is to manipulate it into a separable or linear form, which simplifies the solution process. In our exercise, converting the second order differential equation by substituting \(v = y'\) simplifies it into a first order equation \(v' + v = e^{-t}\). This conversion allows us to easily apply further solving techniques like integration or finding an integrating factor.

Integration

Integration is a fundamental concept in calculus, often described as the reverse process of differentiation. It’s a way to find a function when its derivative is known. The notation for an integral is \(\int f(x) \, dx\), where \(f(x)\) is the integrand. The process of integrating is akin to summing up infinitesimal parts to find the total accumulation or area under the curve represented by a function.
In differential equations, integration is used to solve equations after rewriting them to isolate the derivative of a function. For example, once we find the function for \(v\) in the form \(v = e^{-t}(t + C_1)\), we integrate it to find \(y\). Using the integration by parts technique, we manage to solve for \(y(t)\).

• The integral of \(e^{-t}(t + C_1)\) involves finding the antiderivative.
• With integration by parts, we choose \(u = t + C_1\) and \(dv = e^{-t} \, dt\).

This process builds the bridge of finding the actual function solution for the differential equation presented.

Integrating Factor

The integrating factor is a clever trick used to solve linear first-order differential equations. It's a function, often denoted by \(I(t)\), used to multiply both sides of the equation, facilitating the solution thanks to its properties.
The idea is to transform the left side of the equation into the derivative of a product of two functions. In the differential equation \(v' + v = e^{-t}\), the integrating factor is calculated as \(I(t) = e^{\int 1 \, dt} = e^t\).

• By multiplying the whole equation \(v' + v = e^{-t}\) by \(e^t\), we get \(e^t v' + e^t v = 1\).
• The left-hand side becomes the derivative of \(e^t v\) using the product rule.
• This simplification makes it easy to integrate both sides and find \(v\).

Hence, the integrating factor plays a vital role in simplifying and solving the equation.

Arbitrary Constants

In solving differential equations, particularly involving integration, arbitrary constants often appear. These constants, usually denoted as \(C_1, C_2,\) etc., represent the family of solutions related to the constant integration aspect.
Understanding arbitrary constants is crucial as they account for the initial conditions or specific cases in differential equations. With each integration step in our solution, a new arbitrary constant is introduced. For instance:

• From solving the first order for \(v\), we derive \(C_1\) after integrating.
• In the final integration to solve for \(y(t)\), we introduce \(C_2\).

These constants ensure our solution is not just a particular function but a general one that could be adjusted to fit specific initial conditions or additional constraints.