Question

Solve the initial-value problem for \(x\) as a function of \(t\). \((t+5) \frac{d x}{d t}=x^{2}+1, t>-5, x(1)=\tan 1\)

Short answer

The solution is \(x(t) = \tan(\ln\frac{|t+5|}{6} + 1)\).

Step-by-step solution

Separate the Variables

We begin by separating the variables to solve the differential equation. The given equation is \((t+5) \frac{dx}{dt} = x^2 + 1\). Separate the variables by dividing both sides by \((x^2 + 1)\) and multiplying both sides by \(dt\): \[ \frac{dx}{x^2 + 1} = \frac{dt}{t+5} \]

Integrate Both Sides

Integrate both sides of the equation to find \(x\) as a function of \(t\). The integral of the left-hand side is a standard integral: \[ \int \frac{dx}{x^2 + 1} = \tan^{-1}(x) + C_1 \] where \(C_1\) is the constant of integration. The integral of the right-hand side becomes: \[ \int \frac{dt}{t+5} = \ln|t+5| + C_2 \] where \(C_2\) is another constant of integration.

Solve for x

Equate the two integrals while accounting for the integration constants: \[ \tan^{-1}(x) + C_1 = \ln|t+5| + C_2 \] Solve for \(x\) by subtracting \(C_1\) from each side: \[ \tan^{-1}(x) = \ln|t+5| - C \] where \(C = C_2 - C_1\). Take the tangent of both sides to express \(x\) explicitly: \[ x = \tan(\ln|t+5| - C) \]

Apply the Initial Condition

Use the initial condition \(x(1) = \tan 1\) to find the constant \(C\). Substitute \(t=1\) and \(x = \tan 1\) into the equation: \[ \tan 1 = \tan(\ln|1+5| - C) \] which implies \(\ln 6 - C = 1\). Solve for \(C\): \[ C = \ln 6 - 1 \]

Write the Solution with the Constant

Substitute \(C = \ln 6 - 1\) back into the expression for \(x\): \[ x = \tan(\ln|t+5| - (\ln 6 - 1)) \] which simplifies to \[ x = \tan(\ln|t+5| - \ln 6 + 1) \] Express the function explicitly: \[ x = \tan(\ln\frac{|t+5|}{6} + 1) \]

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In the context of the initial value problem, like the one we have here, differential equations describe the rate of change of a variable. The given differential equation is \[(t+5) \frac{dx}{dt} = x^2 + 1,\] where the objective is to find the function \(x(t)\). The equation involves both \(x\) and its derivative \(\frac{dx}{dt}\), making it a first-order differential equation. Solving these equations gives us insight into how a system evolves over time. In many cases, we seek a solution that also satisfies initial conditions, like \(x(1) = \tan 1\) here, to determine a specific trajectory out of many possible ones.

Variable Separation

Variable separation is a technique used to solve differential equations where the variables can be moved to opposite sides of the equation. In our example, we start with \[(t+5) \frac{dx}{dt} = x^2 + 1.\]The goal is to rearrange the equation so that all terms involving \(x\) are on one side and all terms involving \(t\) are on the other. This is achieved by isolating the differential elements:\[\frac{dx}{x^2 + 1} = \frac{dt}{t+5}.\]This separation allows us to integrate both sides independently, a key step towards solving the differential equation. This method is particularly useful when the equation structure allows it, setting the stage for integration to find particular solutions.

Integrating Factors

Integrating factors are not directly used in the exercise but are important in the context of more complex differential equations. An integrating factor is a function used to multiply a differential equation to simplify it, often transforming a non-exact equation into an exact, solvable one. It is crucial when dealing with linear differential equations that do not easily separate.In the exercise, however, we proceed with direct integration after separating the variables:For instance, \[\int \frac{dx}{x^2 + 1} = \tan^{-1}(x) + C_1,\]and \[\int \frac{dt}{t+5} = \ln|t+5| + C_2.\]These steps lead us to find the general solution before applying initial conditions.

Standard Integrals

Standard integrals are well-known integral results used extensively to solve problems involving calculus, especially differential equations. In our problem, two standard integrals are critical:- The integral of \(\frac{dx}{x^2 + 1}\) is \(\tan^{-1}(x)\).- The integral of \(\frac{dt}{t+5}\) is \(\ln|t+5|\).Using these standard integral results makes solving the differential equation much more straightforward. The selection of these integrals depends on the rearranged form achieved by variable separation. Standard integrals serve as a toolkit for converting complex expressions into simpler forms. Having a strong understanding of these can significantly speed up the process of tackling differential equations, transforming complicated mathematical tasks into more manageable solutions.