Question

Solve the following initial-value problems by using integrating factors. $$ \left(1+x^{2}\right) y^{\prime}=y-1, y(0)=0 $$

Short answer

The solution is \(y(x) = 0\) for all \(x\).

Step-by-step solution

Write the Differential Equation in Standard Form

Start with the given differential equation \((1+x^2) y' = y - 1\). To use the method of integrating factors, we need to express the equation in the standard linear form: \(y' + P(x)y = Q(x)\). First, divide every term by \(1+x^2\) to obtain \(y' - \frac{1}{1+x^2} y = -\frac{1}{1+x^2}\). Now the equation is in standard form with \(P(x) = -\frac{1}{1+x^2}\) and \(Q(x) = -\frac{1}{1+x^2}\).

Determine the Integrating Factor

The integrating factor \(\mu(x)\) is calculated using the formula \(\mu(x) = e^{\int P(x) \, dx}\). Here, \(P(x) = -\frac{1}{1+x^2}\). The integral of \(P(x)\) is \(\int -\frac{1}{1+x^2} \, dx = -\tan^{-1}(x)\). Therefore, the integrating factor is \(\mu(x) = e^{-\tan^{-1}(x)}\), which simplifies by properties of exponents to \(\mu(x) = \frac{1}{e^{\tan^{-1}(x)}}\).

Multiply through by the Integrating Factor

Multiply each term of the standard form equation \(y' - \frac{1}{1+x^2} y = -\frac{1}{1+x^2}\) by the integrating factor \(\mu(x) = \frac{1}{e^{\tan^{-1}(x)}}\). This gives \(\frac{1}{e^{\tan^{-1}(x)}} (y' - \frac{1}{1+x^2} y) = \frac{-1}{(1+x^2)e^{\tan^{-1}(x)}}\).

Right-hand Side Equals a Derivative

The left-hand side of the equation should be recognized as the derivative of \((\mu(x) y)\). Thus, \(\frac{d}{dx}(\frac{y}{e^{\tan^{-1}(x)}}) = \frac{-1}{(1+x^2)e^{\tan^{-1}(x)}}\).

Integrate Both Sides

Integrate both sides with respect to \(x\). \(\int \frac{d}{dx}(\frac{y}{e^{\tan^{-1}(x)}}) \, dx = \int \frac{-1}{(1+x^2)e^{\tan^{-1}(x)}} \, dx\). This simplifies to \(\frac{y}{e^{\tan^{-1}(x)}} = C\), since the right side is difficult to solve analytically in elemental form, we focus on the left to solve it.

Apply the Initial Condition

Use the initial condition \(y(0) = 0\) to find \(C\). Since \(\tan^{-1}(0) = 0\), substitute into the equation: \(\frac{0}{e^{0}} = C\), hence \(C = 0\).

Solve for \(y(x)\)

From \(\frac{y}{e^{\tan^{-1}(x)}} = 0\), solve for \(y\). This results in \(y = 0 \times e^{\tan^{-1}(x)}\) that simplifies to \(y = 0\) for all \(x\).

Key concepts

Initial-Value Problems

Initial-value problems are a type of differential equation that includes conditions specifying the value of the solution at a particular point.
These problems consist of a differential equation along with an initial condition, typically given as \(y(x_0) = y_0\). The initial condition helps in finding the specific solution that passes through a predetermined point on the plane.
In our example, the initial condition is \(y(0) = 0\). This condition ensures that our solution intersects the y-axis at the origin, making it unique.

Initial-value problems are crucial in modeling real-world phenomena where the state of a system is known at a particular time. By solving these, we can determine the system's behavior over time.

Linear Differential Equations

Linear differential equations are equations that involve an unknown function and its derivatives. These equations are linear in terms of the unknown function and its derivatives.
In our specific problem, the differential equation is \((1+x^2) y' = y - 1\), which needs to be rearranged into a standard form.

The standard linear differential form is given by \(y' + P(x)y = Q(x)\). We transform our equation into this form by dividing through by \(1+x^2\), obtaining \(y' - \frac{1}{1+x^2}y = -\frac{1}{1+x^2}\).
Working with linear differential equations allows us to apply various straightforward solving techniques, such as integrating factors, which simplify and solve these equations systematically.

Integrating Factor Derivation

Integrating factors are a powerful tool used to solve linear first-order differential equations. They transform a non-exact equation into an exact one, making it easier to integrate and solve.

To find the integrating factor \(\mu(x)\), we use the formula \(\mu(x) = e^{\int P(x) \, dx}\), where \(P(x)\) is the coefficient of \(y\) in the standard form.
In our problem, after rewriting the equation, \(P(x)\) becomes \(-\frac{1}{1+x^2}\).

• The integral \(\int -\frac{1}{1+x^2} \, dx\) is solved to be \(-\tan^{-1}(x)\).
• The integrating factor is then determined as \(\mu(x) = e^{-\tan^{-1}(x)}\), which simplifies to \(\frac{1}{e^{\tan^{-1}(x)}}\).

By multiplying the entire standard form equation by this integrating factor, we obtain an equation with a left-hand side that is the derivative of \((\mu(x) y)\), allowing for straightforward integration. This process greatly simplifies the task of solving such differential equations.