Problem 19 - Advanced Engineering Mathematics

Question

Given that \(y=-\frac{2}{x}+x\) is a solution of the \(\mathrm{DE} x y^{\prime}+y=2 x\). Find \(x_{0}\) and the largest interval \(I\) for which \(y(x)\) is a solution of the IVP $$ x y^{\prime}+y=2 x, \quad y\left(x_{0}\right)=1 . $$

Step-by-Step Solution

Answer

The value of \(x_0\) is 1, and the largest interval is \((0, \infty)\).

Step 1: Differentiate the Given Function

Start by differentiating the function given in the problem, which is \( y = -\frac{2}{x} + x \). The derivative \( y' \) can be found as follows: for the term \(-\frac{2}{x}\), use the power rule noting it as \(-2x^{-1}\): its derivative is \(2x^{-2}\) or \(\frac{2}{x^2}\). The derivative of \(x\) is simply \(1\). Therefore, \( y' = \frac{2}{x^2} + 1 \).

Step 2: Substitute into the Differential Equation

Substitute \( y \) and \( y' \) into the differential equation \( x y' + y = 2x \). This gives: \[ x \left( \frac{2}{x^2} + 1 \right) + \left( -\frac{2}{x} + x \right) = 2x. \] Simplify the left side: \[ \frac{2}{x} + x - \frac{2}{x} + x = 2x. \] Which simplifies to \( 2x = 2x \), confirming it satisfies the equation.

Step 3: Solve the Initial Value Problem

We need \( y(x_0) = 1 \) under the initial condition for the initial value problem (IVP). Substitute \( x_0 \) into \( y(x) = -\frac{2}{x} + x \) where \( y(x_0) = 1 \): thus, \[ 1 = -\frac{2}{x_0} + x_0. \] Rearrange to find \( x_0 \): \[ x_0^2 - 1 = \frac{2}{x_0}. \] Multiply through by \(x_0\) to clear fractions: \[ x_0^3 - x_0 = 2. \] Solve the cubic equation: in simplified form, we find \( x_0 = 1 \).

Step 4: Determine the Interval of Validity

The function \( y(x) = -\frac{2}{x} + x \) is discontinuous at \( x = 0 \) because of the term \(-\frac{2}{x}\). Thus, the function is valid on the interval \((0, \infty)\). Include the initial condition \( y(1) = 1 \), confirming that \( x_0 = 1 \) is within the interval. Therefore, the largest interval of validity for \( x_0 = 1 \) is \((0, \infty)\).

Key Concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation that comes with a specified initial condition. This condition is usually defined by an equation of the form \( y(x_0) = y_0 \), where \(x_0\) is the specific point at which the initial value is known, and \(y_0\) is the value of the function at that point.
In our exercise, the IVP is given as \( x y^{\prime} + y = 2x \) with the initial condition \( y(x_0) = 1 \). This means that we are looking for a solution to the differential equation that also satisfies \( y(1) = 1 \). This additional piece of information helps us find a specific solution rather than a general one, by fixing the free constant that would typically appear in the solution of a differential equation.

Derivative Calculation

Differentiating functions is a fundamental task when solving differential equations. For the given function \( y = -\frac{2}{x} + x \), we need to find \( y' \), its first derivative.

• The term \(-\frac{2}{x}\) can be rewritten as \(-2x^{-1}\). By applying the power rule, the derivative becomes \(\frac{2}{x^2}\).
• The derivative of the linear term \(x\) is simply \(1\), as the power of \(x\) here is naturally reduced by one (from \(x^1\)).

Therefore, combining these results, the full derivative is \( y' = \frac{2}{x^2} + 1 \). This derivative is then plugged back into the original differential equation to verify if it satisfies the equation, ensuring consistency with the given differential structure.

Interval of Validity

The interval of validity contains all points in the domain of the function for which the solution to the differential equation is both valid and continuous. Discontinuities, such as division by zero, define the boundaries of these intervals.
In the problem, the function \( y(x) = -\frac{2}{x} + x \) contains a term \(-\frac{2}{x}\), which is undefined at \(x = 0\). Therefore, the solution is not valid at this point. As there are no other points of discontinuity, the largest interval where the function remains valid is \((0, \infty)\). This interval covers all positive real numbers, starting right after 0 and extending to infinity, confirming that the initial condition \(x_0 = 1\) falls within this interval.

Solution Verification

To ensure correctness, verifying the solution of the differential equation with the given initial condition is crucial. After determining the derivative \(y'\), substituting both \(y\) and \(y'\) back into the differential equation verifies if they satisfy the equation throughout the interval.

• Substitute \( y = -\frac{2}{x} + x \) and \( y' = \frac{2}{x^2} + 1 \) into \( x y' + y = 2x \).
• Simplify the resulting expression to check equality with the right side \(2x\).

The equation simplifies correctly to \(2x = 2x\), confirming the solution's validity. Additionally, substituting \(x_0 = 1\) into the function \(y(x)\), we solve the initial value correctly, aligning perfectly with \(y(1) = 1\). Therefore, everything checks out, and the solution is certified as correct for the specified conditions and within the interval.