Question

Find the general solution to the differential equation.\(\frac{d y}{d t}=y \cos (3 t+2)\)

Short answer

The general solution is \(y = C' e^{\left(\frac{1}{3} \sin(3t + 2)\right)}\).

Step-by-step solution

Understand the Differential Equation

The given differential equation is \(\frac{dy}{dt} = y \cos(3t + 2)\). It is a first-order, linear differential equation in the form \(\frac{dy}{dt} = f(t)y\), where \(f(t) = \cos(3t + 2)\).

Rearrange and Identify Integrable Form

Rearrange the equation to separate variables: \(\frac{1}{y} dy = \cos(3t + 2) dt\). Our goal is to integrate both sides separately to find the general solution.

Integrate Both Sides

Integrate the left-hand side with respect to \(y\):\[\int \frac{1}{y} \, dy = \ln|y| + C_1\]For the right-hand side, integrate with respect to \(t\):\[\int \cos(3t + 2) \, dt = \frac{1}{3} \sin(3t + 2) + C_2\]

Combine and Simplify Integration Constants

Equate the integrals and combine the constants for simplicity:\[\ln|y| = \frac{1}{3} \sin(3t + 2) + C\]where \(C = C_2 - C_1\).

Solve for y

To express \(y\) explicitly, exponentiate both sides to remove the natural logarithm:\[y = e^{\left(\frac{1}{3} \sin(3t + 2) + C\right)} = e^C \cdot e^{\left(\frac{1}{3} \sin(3t + 2)\right)}\]Let \(C' = e^C\), then the general solution is:\[y = C' e^{\left(\frac{1}{3} \sin(3t + 2)\right)}\]

Key concepts

Separation of Variables

Separation of variables is a fundamental method to solve ordinary differential equations, especially first-order ones. This approach involves rearranging the equation such that all terms containing one variable, in our case either dependent or independent, are on one side of the equation, and the terms containing the other variable are on the other side. This makes it possible to integrate each side separately, paving the way to finding a solution.

In the example provided, \[\frac{dy}{dt} = y \cos(3t + 2)\]we start by dividing both sides by \(y\) and multiplying by \(dt\) to get:\[\frac{1}{y} dy = \cos(3t + 2) dt\]

This is the step where separation of variables is successfully applied. It transforms the original problem allowing the integration of each side independently, aiding in bridging the differential equation to its eventual solution.

First-Order Linear Differential Equation

A first-order linear differential equation is any equation of the form \[\frac{dy}{dt} = f(t)y + g(t)\]. In such equations, the highest derivative is the first one. They appear frequently in mathematical modeling of natural phenomena, like population growth, radioactive decay, and circuit dynamics.

In our given exercise,\[\frac{dy}{dt} = y \cos(3t + 2)\] can be interpreted as linear because it matches the form \(\frac{dy}{dt} = f(t)y\), where \(f(t) = \cos(3t + 2)\). There is no \(g(t)\) term here which would appear on its own, not multiplied by \(y\). Such linear equations exhibit the key property that their general solution involves integrating a function of \(t\), making them straightforward to tackle using separation of variables.

To solve these, you typically transform the equation in a manner facilitating straightforward integration, often involving natural logarithms when integrating the \(dy/y\) term, as seen in the solution process above.

Integration of Trigonometric Functions

Trigonometric functions frequently crop up in differential equations and integration, knowing how to handle their integrals is invaluable. Basic trigonometric integrals are foundational knowledge:

• \(\int \sin(at) \, dt = -\frac{1}{a} \cos(at) + C\)
• \(\int \cos(at) \, dt = \frac{1}{a} \sin(at) + C\)

In our specific solution, the integration required is:
\[\int \cos(3t + 2) \, dt = \frac{1}{3} \sin(3t + 2) + C_2\]

The factor of \(\frac{1}{3}\) comes from the chain rule, where the derivative of \((3t+2)\) with respect to \(t\) is \(3\). Recognizing this allows us to account for the constant multiplier that appears when performing this integral, leading to a proper solution. This understanding is critical for solving many problems involving applied calculus or physics. Overall, being adept at integrating trigonometric functions equips you to tackle a wide array of integrals you may encounter.