Question

For the following problems, find the general solution to the differential equation.\(x^{\prime}=\operatorname{coth} t+\ln t+3 t^{2}\)

Short answer

The general solution is \( x(t) = \ln|\sinh t| + t \ln t - t + t^3 + C \).

Step-by-step solution

Understand the Differential Equation

The given differential equation is \( x' = \coth t + \ln t + 3t^2 \). This is a first-order ordinary differential equation (ODE). Essentially, we need to integrate the right-hand side to find the general solution for \( x(t) \).

Integrate the Hyperbolic Cotangent

We begin integrating the first term \( \coth t \). Recall that \( \coth t = \frac{\cosh t}{\sinh t} \), and its integral is \( \ln|\sinh t| + C_1 \), where \( C_1 \) is the constant of integration.

Integrate the Natural Logarithm

Next, integrate the second term \( \ln t \). Use integration by parts where \( u = \ln t \) and \( dv = dt \). Then, \( du = \frac{1}{t} dt \) and \( v = t \) giving us \( t \ln t - \int t \cdot \frac{1}{t} dt = t \ln t - t + C_2 \).

Integrate the Polynomial Term

Finally, integrate the polynomial term \( 3t^2 \). The integral of \( 3t^2 \) is \( t^3 + C_3 \), where \( C_3 \) is another constant of integration.

Combine Integrals for General Solution

Combine all the integrals found in the previous steps to get the general solution:\[ x(t) = \ln|\sinh t| + t \ln t - t + t^3 + C \]where \( C = C_1 + C_2 + C_3 \) is the overall constant of integration.

Key concepts

Integration

Integration is a fundamental concept in calculus, acting as a reverse operation to differentiation. It helps us find functions when their derivatives are known, or evaluate areas under curves.
To "integrate" means to find an antiderivative or an integral of a function. In other words, it's a way to "add up" an infinite number of infinitesimal quantities.There are two main types of integration:

• Indefinite Integration: Finding the general form of antiderivative, represented with an integration constant (e.g., \( \int f(x) \, dx = F(x) + C \)).
• Definite Integration: Calculating the exact area under the curve of a function between two points, which produces a numerical result.

In tackling differential equations, integration allows us to accumulate solutions to unknown functions based on their rate of change, which is often represented by the derivative.
This is how we solve ordinary differential equations like the one we saw, helping us combine multiple antiderivative calculations into a single, comprehensive solution.

Hyperbolic Functions

Hyperbolic functions mirror the trigonometric functions with some unique properties and play a significant role in various calculus problems. Functions such as hyperbolic sine, cosine, and tangent, noted as \( \sinh t \), \( \cosh t \), and \( \tanh t \) respectively, are defined through exponential functions.One important hyperbolic function is the hyperbolic cotangent, \( \coth t \), defined as:\[\coth t = \frac{\cosh t}{\sinh t}\]It behaves similarly to its trigonometric counterpart, cotangent.When solving differential equations involving hyperbolic functions, it's important to recognize their derivatives and integrals:

• \( \frac{d}{dt}\coth t = - \operatorname{csch}^2 t \)
• \( \int \coth t \; dt = \ln|\sinh t| + C \)

Understanding these properties allows the solver to rewrite and manipulate the differential equations,effectively finding solutions that incorporate hyperbolic terms.

Integration by Parts

Integration by parts is a technique in calculus, derived from the product rule of differentiation. It is particularly useful when dealing with integrals of products of functions.
This method transforms an "unfriendly" integral into a more approachable form:The integration by parts formula is given by:\[\int u \, dv = uv - \int v \, du\]Where:

• \( u \) and \( dv \) are parts of the integrand.
• \( du \) and \( v \) are their respective derivations and antiderivatives.

For example, solving the integral of \( \ln t \) in our equation is simplified using integration by parts:- Let \( u = \ln t \); therefore, \( du = \frac{1}{t} dt \).- Let \( dv = dt \); therefore, \( v = t \).Applying the formula gives:\[t \ln t - \int t \cdot \frac{1}{t} dt = t \ln t - t + C_2\]Thus, this technique is immensely helpful when tackling complex integrals in differential equations, allowing for breaking down problems into simpler components.