Question

Using the given boundary condition, find the particular solution to each differential equation. Try some by calculator. $$x(y+1) y^{\prime}=y(1+x), x=1 \text { when } y=1$$

Short answer

The particular solution to the differential equation is \(y = Ce^{2x}x^3\), where C = 6/(2^3e^4).

Step-by-step solution

Separate the Variables

In order to solve this first order differential equation, separate the variables to one side of the equation related to y and the other related to x. The given differential equation is \(y'=2+\frac{3y}{x}\). To separate, re-write the equation as \(y' - \frac{3y}{x} = 2\), and arrange it to the form \(\frac{dy}{dx} = \frac{3y}{x} + 2\).

Integrate Both Sides

After separating, integrate both sides with respect to their own variables to find the general solution. The integral form will be \(\int \frac{1}{y} dy = 3\int \frac{1}{x} dx + \int 2dx\).

Find the General Solution

Upon integrating both sides, the equation becomes \(\ln|y| = 3\ln|x| + 2x + C\), where C is the constant of integration.

Solve for the Particular Solution Using the Initial Condition

Use the boundary condition \(x=2\) and \(y=6\) to solve for the constant C and find the particular solution. Plugging in the values gives us \(\ln|6| = 3\ln|2| + 2(2) + C\). Solve for C and then substitute back into the equation to get the particular solution.

Express the Particular Solution Explicitly

After solving for C, you will use the relationship \(e^{\ln|y|} = y\) to rewrite the solution explicitly in terms of y by exponentiating both sides of the equation.

Key concepts

Separation of Variables

In the realm of first order differential equations, separation of variables is a method often employed as a first resort. The technique involves rearranging the equation to isolate each variable with its respective derivative. Think of it as a sorting task, where you're grouping all the 'y' associated terms on one side, and those with 'x' on the other.

For the given equation
\[ y' = 2 + \frac{3y}{x} \]
the process would look something like subtracting \( \frac{3y}{x} \) from both sides to get
\[ y' - \frac{3y}{x} = 2 \]
and then further rearranging to have 'dy' and 'dx' separated, leading to
\[ \frac{dy}{dx} = \frac{3y}{x} + 2 \]
Once in this form, we're able to perform integration on each side, setting the stage for finding a general solution.

Integrating Differential Equations

Now that we’ve separated our variables, it's time to integrate. Integration is the reverse of differentiation, essentially 'un-doing' the derivative to find the original function (or in our case, the general solution).

Integrating each side of
\[ \frac{dy}{dx} = \frac{3y}{x} + 2 \]
involves finding the antiderivative of each side. With
\[ \int \frac{1}{y} dy = 3\int \frac{1}{x} dx + \int 2dx \]
you'll arrive at
\[ \ln|y| = 3\ln|x| + 2x + C \]
where 'C' is the constant of integration. This represents all of the infinite vertically shifted forms of our original function, which is why our next step will be to pin down the specific case we're dealing with by applying an initial condition.

Initial Conditions

You have the general solution, but it's like having a map with endless paths -- initial conditions help us find the one path that leads to our destination, or the particular solution. In math, an initial condition is a specific point on the graph of the solution, which in our case is given by
\[ x=2 \]
when
\[ y=6 \]
When you plug these values into the general solution
\[ \ln|y| = 3\ln|x| + 2x + C \]
you're able to solve for 'C'. It’s a bit like solving a mystery—the initial conditions give you the clue to unveil the secret identity of 'C', and consequently, the particular solution.

Particular Solution

The particular solution is the tailored fit to our differential equation, honoring the given initial conditions. To find it, we use the value of 'C' obtained in the previous step and substitute it back into our general solution equation. This process refines the infinite possibilities down to the one equation that passes through the point given by our initial conditions.

After finding 'C', you express the final answer explicitly in terms of 'y':
\[ e^{\ln|y|} = y \]
allowing us to write the particular solution in a neatly packaged form that includes the constant and makes the relationship between 'x' and 'y' clear.