Question

Using the given boundary condition, find the particular solution to each differential equation. $$5 x=2 y^{2}+4 x y y^{\prime}, x=1 \text { when } y=4$$

Short answer

The particular solution can be found by solving for y' using the given boundary condition and integrating if needed.

Step-by-step solution

Write down the differential equation

First, we recognize the given differential equation which is in implicit form: 5x = 2y^2 + 4xyy'.

Differentiate implicitly with respect to x

We need to find the derivative of both sides of the equation with respect to x which involves using implicit differentiation, keeping in mind that y is a function of x. The derivative of 5x with respect to x is 5. The derivative of 2y^2 with respect to x is 4yy' because of the chain rule. Finally, the derivative of 4xyy' with respect to x is 4y' + 4xy'', by using the product rule on 4xy and then multiplying by y'.

Substitute the initial condition into the equation

We are given that x = 1 when y = 4. Substituting these values into our differentiated equation allows us to solve for the value of y'.

Solve for y'

After substituting the initial conditions into the differentiated equation, we can isolate y' to find its value. This gives us the slope of the solution curve at the point where x = 1 and y = 4.

Find the particular solution

At this point, we should have an expression for y' involving y and x. We use this, along with the given boundary conditions, to solve for the particular solution to the differential equation.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with one or more of its derivatives. They are used to model a vast array of phenomena in science and engineering, such as the motion of objects under forces, the flow of current in electrical circuits, and the change of populations over time. The key to solving a differential equation is to understand the relationship between the variables and their rates of change.

In our exercise, we encounter an implicit differential equation, where the relationship between the dependent variable, usually denoted as 'y', and the independent variable 'x', is not given explicitly. Instead, the equation provides a condition that the solution must satisfy. In this case, the differential equation describes a geometric relationship between the variables and their derivatives, and can often represent complex physical situations.

Boundary Conditions

Boundary conditions are essential in determining a unique solution to a differential equation. In real-life problems, these conditions often correspond to known states or behaviors of a system at specific points. For example, the initial temperature at various points on a metal rod, or the initial position and velocity of a moving particle.

In our problem, the boundary condition is given as x = 1 when y = 4. This specific type of boundary condition is an initial condition, which tells us the starting state of the system. The boundary condition serves as a 'seed' from which the particular solution to the differential equation will grow. This allows us to find the specific solution that not only satisfies the differential equation but also the given condition.

Chain Rule

The chain rule is an essential technique in calculus for taking the derivative of composite functions, which are functions composed of other functions. When we have a function of a function, like y(x), and we want to differentiate it with respect to x, the chain rule comes into play.

In the context of our exercise, when differentiating a term like \(2y^2\), where y is a function of x, we consider the outer function \(u^2\) and the inner function u(y), which is actually y(x). Applying the chain rule, we differentiate \(u^2\) to get \(2u\) and multiply by the derivative of the inner function, which is y'. This process is central to implicit differentiation, which is used when y cannot be easily isolated on one side of the equation.

Product Rule

The product rule is used in calculus for differentiating products of two or more functions. It states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. In formula form, if we have functions u(x) and v(x), then \[ (u \times v)' = u' \times v + u \times v'. \]

In the problem, we apply the product rule to differentiate the term \(4xyy'\). Since 4xy is a product of two functions, 4x and y, and y itself is a function of x, we need the product rule to take the derivative correctly. It allows us to calculate the derivative piece by piece and then reassemble the parts, yielding the term \(4y' + 4xy''\) as required.