Question

Finding general solutions Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots\) to denote arbitrary constants. $$y^{\prime}(t)=t \ln t+1$$

Short answer

Answer: The general solution is \(y(t) = \ln t\cdot\frac{t^2}{2} - \frac{t^2}{4} + t + C\), where \(C\) is an arbitrary constant.

Step-by-step solution

Find the antiderivative of the first term, \(t\ln t\)

To find the antiderivative of \(t\ln t\), we will use integration by parts. Integration by parts states: $$\int u \, dv = uv - \int v \, du$$ We let \(u = \ln t\) and \(dv = t\,dt\). Differentiating \(u\) with respect to \(t\), we get \(du = \frac{1}{t}\,dt\). Integrating \(dv\) with respect to \(t\), we have \(v = \frac{t^2}{2}\). Now we plug these into the integration by parts formula: $$\int t\ln t \, dt = \int u \, dv = uv - \int v \, du = \ln t\cdot \frac{t^2}{2}-\int \frac{t^2}{2}\cdot\frac{1}{t}\,dt$$

Simplify the integral in the expression

We can simplify the remaining integral as such: $$\int \frac{t^2}{2}\cdot\frac{1}{t}\,dt = \int \frac{t}{2}\, dt$$

Integrate the simplified expression

Now we can integrate the simplified expression: $$\int \frac{t}{2}\, dt = \frac{1}{2}\int t\, dt = \frac{1}{2}\cdot\frac{t^2}{2}+C_1 = \frac{t^2}{4} + C_1$$

Substitute the result back into the integration by parts expression

Now we can substitute back the result of the simplified integral to the integration by parts expression: $$ \int t\ln t\, dt = \ln t \cdot \frac{t^2}{2} - \left(\frac{t^2}{4} + C_1\right) $$

Integrate the second term, 1

Now, we need to find the antiderivative of the second term, which is \(1\): $$ \int 1\, dt = t + C_2 $$

Combine the antiderivatives to find the general solution

Finally, we will combine the antiderivatives to find the general solution of the differential equation: $$ y(t) = \left(\ln t \cdot \frac{t^2}{2} -\frac{t^2}{4}\right) + \left(t + C_2\right) + C_1 $$ We can combine the constants \(C_1\) and \(C_2\) into a single constant \(C\): $$ y(t) = \ln t\cdot\frac{t^2}{2} - \frac{t^2}{4} + t + C $$ This is the general solution to the given differential equation.

Key concepts

Integration by Parts

Understanding integration by parts is essential when solving certain types of integrals, particularly those involving products of functions like polynomials and logarithms. The formula for integration by parts comes from the product rule for differentiation and is given by \[\int u \, dv = uv - \int v \, du\]
In the process, we pick functions for \(u\) and \(dv\) such that the derivative of \(u\) and the integral of \(dv\) are simpler to work with. A common strategy for selecting \(u\) is the LIATE rule, which suggests picking logarithmic, inverse trigonometric, algebraic, trigonometric, or exponential functions, in that order of preference.

• Choose \(u\): Pick a function that becomes simpler upon differentiation.
• Choose \(dv\): Select the remaining function, which should be easy to integrate.
• Compute \(du\) and \(v\): Take the derivative of \(u\) and the antiderivative of \(dv\).
• Apply the formula: Substitute everything into the integration by parts formula.

When applying integration by parts to our exercise, we select \(u = \ln t\) and \(dv = t\,dt\), because the derivative of \(\ln t\) gets rid of the logarithm, while the integral of \(t\) is straightforward. This makes it a prime candidate for integration by parts and leads us smoothly to the antiderivative of \(t\ln t\).

Antiderivative

An antiderivative, or indefinite integral, of a function \(f(t)\) is a differentiable function \(F(t)\) whose derivative is \(f(t)\). Essentially, finding an antiderivative means reversing the process of differentiation. The general form of an antiderivative is written as\[F(t) = \int f(t)\, dt\]
and the arbitrary constant of integration \(C\) is added to account for all possible vertical shifts of the antiderivative function. Every antiderivative has infinitely many forms, differing only by this constant of integration.

Practical Tips:

• Understand common antiderivatives like the power rule \(\int t^n dt = \frac{t^{n+1}}{n+1} + C\), when \(n\) is not equal to -1.
• Recognize the antiderivatives of trigonometric functions and exponential functions.
• Use substitution when the function is a composition of a known function and its derivative.

In the step by step solution, after applying integration by parts, the antiderivative of the simpler expression \(\int \frac{t}{2} dt\) is found using the power rule. Similarly, the integral of the constant function \(1\) is just \(t\), plus the constant of integration \(C_2\). Combining these results provides us with the general solution of the differential equation.

Differential Equation

A differential equation is an equation that relates a function with its derivatives. It defines the dynamic relationship between quantities and how they change with respect to change in another quantity, often time. Solving a differential equation typically means finding the general solution, which includes arbitrary constants to account for all initial conditions.

Types of Differential Equations:

• Ordinary Differential Equations (ODEs): Equations containing functions of only one independent variable and its derivatives.
• Partial Differential Equations (PDEs): Equations involving partial derivatives of functions with more than one independent variable.

The differential equation in our exercise, \(y'(t) = t \ln t + 1\), is an ODE where \(y'(t)\) represents the derivative of \(y\) with respect to \(t\). The goal is to integrate both sides with respect to \(t\) to find \(y(t)\), the function defining the behavior of the system described by the equation, incorporating the initial conditions through the constant of integration \(C\). Understanding the nature of differential equations and their solutions is fundamental in fields such as physics, engineering, and economics, where they model a vast array of physical phenomena and dynamic systems.