Question

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \(y^{\prime}-x y=x\) \(y=1\) when \(x=0\)

Short answer

The particular solution is \( y = 2 e^{\frac{x^2}{2}} - 1 \).

Step-by-step solution

Rewrite the Differential Equation

Rewrite the given differential equation to facilitate separation of variables.The given differential equation is:\[ y^{\text{'}'} - xy = x \]Rewriting it, we get:\[ \frac{dy}{dx} - xy = x \]

Separate the Variables

Separate the variables so that all terms involving y are on one side of the equation and all terms involving x are on the other side.Adding xy to both sides, we have:\[ \frac{dy}{dx} = xy + x \]Factor out the x on the right-hand side:\[ \frac{dy}{dx} = x(y + 1) \]Divide both sides by y + 1:\[ \frac{1}{y+1} \frac{dy}{dx} = x \]

Integrate Both Sides

Integrate both sides with respect to x to find the general solution.\[ \frac{1}{y+1} dy = x dx \]Integrate both sides:\[ \begin{aligned} \int \frac{1}{y+1} dy &= \int x \, dx \ \ln|y+1| &= \frac{x^2}{2} + C \ \ y + 1 &= e^{\frac{x^2}{2} + C} \end{aligned} \]Exponentiating both sides, we get:\[ y + 1 = e^{\frac{x^2}{2}} e^C \]Letting \( e^C = K \), we obtain the general solution:\[ y + 1 = K e^{\frac{x^2}{2}} \]Solving for y, we get:\[ y = K e^{\frac{x^2}{2}} - 1\]

Apply Boundary Condition

Use the given boundary condition, \( y(0) = 1 \), to find the particular solution.Since \( y = 1 \) when \( x = 0 \):\[ 1 = K e^{\frac{0^2}{2}} - 1 \]\[ 1 = K - 1 \]\[ K = 2 \]Substitute K back into the general solution:\[ y = 2 e^{\frac{x^2}{2}} - 1 \]

Key concepts

Separation of Variables

Separation of variables is a method used to solve ordinary differential equations (ODEs). The main idea is to move all terms involving one variable to one side of the equation and all terms involving the other variable to the opposite side. This way, you can integrate both sides separately.

Let's take the given differential equation: \(\frac{dy}{dx} - xy = x\) and rearrange it to make it easier to separate the variables. By adding \(xy\) to both sides, we get: \(\frac{dy}{dx} = x(y + 1)\).

Now, we divide both sides by \(y + 1\) to isolate the terms involving \y\ on one side: \(\frac{1}{y + 1} \frac{dy}{dx} = x\). Notice that each side now has only one variable, which allows us to integrate both sides independently.

General Solution

The general solution of a differential equation contains an arbitrary constant, denoted by \(C\), which represents an infinite number of possible solutions. In terms of our example:

Once we have separated the variables: \( \frac{1}{y + 1} dy = x dx \), we integrate both sides:

\(\begin{aligned} \int \frac{1}{y + 1} dy &= \int x \, dx \ \ln|y + 1| &= \frac{x^2}{2} + C \ y + 1 &= e^{\frac{x^2}{2} + C} \ \text{Let } e^C &= K \ y + 1 &= K e^{\frac{x^2}{2}} \ y &= K e^{\frac{x^2}{2}} - 1 \end{aligned} \)

This equation defines the general solution because it incorporates the constant \(K\), allowing for a wide range of solutions depending on the value of \(K\).

Particular Solution

To find a particular solution, we must use additional conditions provided in the problem, such as initial conditions or boundary conditions. This specific solution will not contain any arbitrary constants.

In this problem, we are given the boundary condition \(y(0) = 1\). Plugging in these values into the general solution \(y = K e^{\frac{x^2}{2}} - 1\), we get:

\( 1 = K e^{\frac{0^2}{2}} - 1 \ 1 = K - 1 \ K = 2 \)

Substituting \(K\) back into the general solution, we find the particular solution:

\( y = 2 e^{\frac{x^2}{2}} - 1 \).

So, \(y = 2 e^{\frac{x^2}{2}} - 1\) is the specific solution that satisfies both the differential equation and the given boundary condition.

Boundary Conditions

Boundary conditions are constraints needed to find a specific solution to a differential equation from the general solution. They provide necessary values at specific points, often given as \(y(x)\) at \(x = a\).

In the example problem, the boundary condition is \(y(0) = 1\). This information tells us that when \(x = 0\), \(y\) should be 1. We use this constraint to determine the value of the arbitrary constant in the general solution.

By substituting \(x = 0\) and \(y = 1\) into the general solution \(y = K e^{\frac{x^2}{2}} - 1\), we solve for \(K\):

\( 1 = K e^{\frac{0^2}{2}} - 1 \ 1 = K - 1 \ K = 2 \)

This provides a unique solution that adheres to the given boundary condition. Boundary conditions are essential because they transform a general solution into a particular solution, making it specific to the problem at hand.