Question

$$ \left(e^{v}+1\right) \cos u d u+e^{v}(\sin u+1) d v=0 $$

Short answer

Rewrite and rearrange the given equation as: \(\frac{\cos u}{\sin u + 1} \ du = -e^{-v} \ dv\). Integrate both sides, perform the substitution: \(t = \sin u + 1\). The final solution is: \[ \ln|\sin u + 1| = -e^{-v} + C \]

Step-by-step solution

Rewrite the equation for better clarity

Rewrite the given equation as follows: \[ (e^v + 1)\cos u \ du + e^v(\sin u+1) \ dv = 0 \]

Rearrange terms to isolate du and dv

Rearrange the equation to isolate du and dv on opposite sides: \[ \frac{\cos u}{\sin u + 1} \ du = -e^{-v} \ dv \]

Integrate both sides

Now, we can integrate both sides of the equation to find the relation between u and v: \[ \int \frac{\cos u}{\sin u + 1} \ du = \int -e^{-v} \ dv \]

Performing the integration

Integrate both sides. To do the left-side integration, perform the substitution: let \(t = \sin u + 1\), so \(\frac{dt}{du} = \cos u\). Then, \(dt = \cos u \ du\). \[ \int \frac{1}{t} \ dt = -\int e^{-v} \ dv \] Upon integrating, we get: \[ \ln|t| = -e^{-v} + C \]

Substitute back and simplify

Substitute back the original expression for t and rewrite the equation in terms of u and v: \[ \ln|\sin u + 1| = -e^{-v} + C \] So, this is the solution for the given differential equation.

Key concepts

Integrating Factors

Integrating factors are a powerful tool used to solve certain types of linear ordinary differential equations (ODEs). When working with a first-order linear ODE of the form \[ \frac{dy}{dx} + P(x)y = Q(x) \], if the equation is not already integrable, an integrating factor can make it so. An integrating factor is essentially a function, usually denoted by \(\mu(x)\), that when multiplied by the original equation, allows the left side to be expressed as the derivative of a product of functions.

The integrating factor is typically chosen such that \(\mu(x)\) solves the equation \[ \frac{d\mu}{dx} = P(x)\mu \.\] Once \(\mu(x)\) is determined, you multiply it through the differential equation, resulting in an equation that is straightforward to integrate.Finding an integrating factor requires knowledge of integration techniques and can greatly simplify the solution process for complex differential equations.

However, in the exercise provided, the differential equation does not explicitly require an integrating factor for its solution. Instead, the problem was solved using separation of variables – another effective method for solving ODEs – but it's important to recognize that integrating factors can be a valuable alternative in different circumstances.

Separable Differential Equations

Separable differential equations are a subset of ordinary differential equations where the variables can be separated on either side of the equation. This technique allows us to take an equation of the form \(\frac{dy}{dx}=g(x)h(y)\) and arrange it such that all terms involving \(x\) are on one side of the equation and all terms involving \(y\) are on the other side, typically resulting in an expression like \(\frac{1}{h(y)}dy=g(x)dx\).

Once the variables are separated, you can integrate both sides independently to find the solution to the differential equation. The given exercise demonstrates this method effectively. After rearranging the given ODE, \( \cos u \) terms are grouped on one side and \( e^{-v} \) terms on the other, enabling separate integration of \(u\) and \(v\) related terms. The separable nature simplifies the originally complex problem, highlighting the elegance and practicality of this approach.

Integration Techniques

Integration techniques are an essential part of solving differential equations and are critical for determining the solutions of many calculus problems. To solve the exercise at hand, specifically step 4 of the solution, we employed substitution, one of the many techniques in integration. Substitution is used when an integral contains a composite function, and involves replacing a part of the integrand with a new variable to simplify the problem.

In our case, \(t = \sin u + 1\) was let and \(\frac{dt}{du} = \cos u\) was deduced, which simplified the integration process. Some other common integration techniques include integration by parts, partial fractions, trigonometric substitution, and using integration tables. Mastery of these techniques is crucial because they enable us to tackle a wide variety of integrals, leading to the solution of more complex differential equations.

These techniques can transform an intractable integral into a manageable one, and as such, the ability to select the appropriate technique for a given problem is a powerful skill in any mathematician's toolkit.