Question

In Exercises \(11-20,\) solve the initial value problem explicitly. \(\frac{d y}{d t}=\frac{1}{1+t^{2}}+2^{t} \ln 2\) and \(y=3\) when \(t=0\)

Short answer

The solution to the given initial value problem is :\(y = \arctan(t) + \frac{2^{t+1}}{ \ln 2} + C\) where \(C\) is determined by the initial conditions. Specifically, for this problem where \(y(0) = 3\), \(C\) has a particular value to make this condition true.

Step-by-step solution

Separate the Equation

Start by separating the equation \(\frac{d y}{d t} =\frac{1}{1+t^{2}}+2^{t} \ln 2 \). We obtain: \(\frac{d y}{d t} - 2^{t} \ln 2=\frac{1}{1+t^{2}}\)

Integrate Both Sides

Next, integrate both sides of the equation with respect to their independent variables. For the left side, we obtain: \[y -\frac{2^{t+1}}{ \ln 2}\]. And on the right side, \[\int \frac{1}{1+t^{2}} dt = \arctan(t)\]. Now, you add a constant of integration \(C\) on one of the two sides.

Apply the Initial Condition

Apply the initial condition \(y=3\) when \(t=0\). So, we have \[3 -\frac{2^{0+1}}{ \ln 2}= \arctan(0) + C\]. Simplify equation to find the value of \(C\).

Write The Final Solution

The strategy is to write down the solution of the original differential equation. Replace \(C\) from step 3 into the result from step 2 to get the complete solution for \(y\).

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are fundamental in expressing the laws of nature and engineering, encapsulating how physical, biological, or economic systems evolve over time.

At their core, differential equations represent the rate of change of some quantity. For instance, they can model how a population grows, how heat dissipates, or how a spring oscillates. The equation from our initial value problem, \( \frac{d y}{d t}=\frac{1}{1+t^{2}}+2^{t} \ln 2 \), is an example of a first-order differential equation since it involves the first derivative of the variable \(y\) with respect to \(t\).

Solving a differential equation often means finding a function that satisfies the relationship defined by the equation. In this case, we look for a function \(y\) that changes over time \(t\) in a manner consistent with the equation provided. The process involves integration, which leads us to the next important topic: integrating factors.

Integrating Factors

An integrating factor is a function used to simplify a differential equation, making it easier to solve. Although not needed in every differential equation, integrating factors are particularly useful when dealing with linear first-order differential equations, which can't be easily separated into functions of \(y\) and \(t\) alone.

In the solution to our initial value problem, the term \(2^t \ln 2\) complicates the direct integration of both sides. However, because our equation lacks this term on one side, we are fortunate not to require an integrating factor for this particular problem. Nevertheless, the concept of an integrating factor is crucial for more complex equations where direct separation is not possible.

Typically, an integrating factor is a function of \(t\) (or whatever the independent variable is) and once found, is multiplied across the entire differential equation to allow terms to combine in a way that facilitatesthe integration of both sides.

Separation of Variables

Separation of variables is a method for solving differential equations where you can separate the variables involved into two sides of the equation - one side containing only \(y\) and \(dy\), and the other containing only \(t\) and \(dt\). It's particularly handy for first-order, separable differential equations as we saw in our initial value problem.

In Step 1 of our solution, we separated the variables by moving the terms involving \(t\) to one side and the differential \(dy\) to the other side. This transformation allows us to integrate the equation with respect to each variable independently, which we performed in Step 2.

The beauty of separation of variables lies in its simplicity. This technique does not always apply to every differential equation, especially those that can't be separated into single-variable expressions. However, when valid, it provides a straightforward path to finding the general solution to a differential equation before applying an initial condition to solve for any constants—leading to the explicit solution we seek.