Question

Finding a Particular Solution Using Separation of Variables In Exercises \(15-24\) , find the particular solution that satisfies the initial condition. $$ \frac{d u}{d v}=u v \sin v^{2} \quad u(0)=1 $$

Short answer

The solution of the differential equation is \(u = e^{-\frac{1}{2}cos(v^2)}\).

Step-by-step solution

Separate variables

Rewrite the given differential equation in a form where all 'u' terms and its differential are on one side and all 'v' terms and its differential are on the other. This gives \(\frac{1}{u}du = v \sin(v^2) dv\).

Substitute to simplify integration

Let \(w = v^2\). This means that \(dw = 2vdv \rightarrow dv = \frac{dw}{2v}\). Substituting \(w = v^2\) and the value for \(dv\) to the equation gives: \(\frac{1}{u}du = \frac{1}{2}sin(w) dw\).

Carry out the Integrations

Integrating both sides yields: \(\int\frac{1}{u}du = \int\frac{1}{2}sin(w) dw\) -> \(ln|u| = -\frac{1}{2}cos(w) + C\), where C is the constant of integration. To clear the logarithm, exponentiate both sides of the equation to yield \(u = e^{-\frac{1}{2}cos(w) + C}\). Now substitute back \(w = v^2\) to get: \(u = e^{-\frac{1}{2}cos(v^2) + C}\). Additionally, to express \(u\) in an easier form, you can write it as \(u = e^C e^{-\frac{1}{2}cos(v^2)}\). Now, let \(A = e^C\), which is yet another constant, to get: \(u = A e^{-\frac{1}{2}cos(v^2)}\).

Apply Initial Condition

Now, apply the initial condition \(u(0)=1\) by evaluating \(u\) at \(v=0\) to find the value of constant A. It yields \(1 = A*e^0\) since \(cos(0) = 1\). Hence, \(A = 1\).

Write final solution

Substitute \(A = 1\) to get the final solution \(u = e^{-\frac{1}{2}cos(v^2)}\).

Key concepts

Differential Equations

Differential equations are equations involving derivatives of a function. They describe how a particular phenomenon evolves over time, such as population growth, heat dissipation, or wave propagation. In this problem, the given differential equation is \( \frac{du}{dv} = uv \sin v^2 \). This represents the rate of change of \( u \) with respect to \( v \).

Understanding differential equations allows us to model and predict behavior in various scientific fields. To solve them, we employ different methods, one of which is the separation of variables. This method is applicable when we can easily split the equation such that all terms involving one variable and its differentiation are on one side, and the other variables are on the opposite side.

Initial Condition

Initial conditions are specific values that allow us to find a particular solution to a differential equation. They give us the value of the function at a specific point. In our problem, the initial condition is \( u(0) = 1 \).

These conditions are crucial because differential equations can have infinitely many solutions. The initial condition uniquely determines one solution based on the given information.

• They guide us in finding specific values of constants that appear during integration.
• They ensure the solution is applicable to the real-world scenario we are modeling.

Applying the initial condition involves substituting the given values into the solution formula and solving for any constants.

Integrating Factor

The integrating factor is a technique often used to solve linear differential equations. However, for this exercise, we employed a different method due to the structure of the equation. While the integrating factor is typically used to convert a non-exact differential equation into an exact one, separation of variables was more appropriate here.

When using separation of variables, we rearrange the equation so that each side pertains to a single variable. This approach then allows us to integrate each side independently. The integration step often introduces a constant, which we later solve using our initial condition.

Particular Solution

The particular solution to a differential equation is the specific solution that satisfies the given initial conditions. It offers a unique solution from the general solution by determining the constants introduced during integration.

In our problem, after applying the initial condition \( u(0) = 1 \), we found the particular solution \( u = e^{-\frac{1}{2}\cos(v^2)} \).

• The particular solution gives us a specific function that describes real-world behavior according to the model.
• It ties together the problem's conditions, offering a precise prediction or description.

These solutions are valuable in practice because they form the foundation for further analysis or forecasting in various scientific and engineering tasks.