Problem 19 - Advanced Engineering Mathematics

Question

(a) Without solving, explain why the initial-value problem $$ \frac{d y}{d x}=\sqrt{y}, \quad y\left(x_{0}\right)=y_{0}, $$ has no solution for \(y_{0}<0\). (b) Solve the initial-value problem in part (a) for \(y_{0}>0\) and find the largest interval \(I\) on which the solution is defined.

Step-by-Step Solution

Answer

For \( y_0 < 0 \), no solution exists. For \( y_0 > 0 \), the solution is \( y = \left( \frac{x - x_0 + 2\sqrt{y_0}}{2} \right)^2 \) on \( [x_0 - 2\sqrt{y_0}, \infty) \).

Step 1: Understanding Why There Is No Solution for y_0 < 0

The differential equation \( \frac{dy}{dx} = \sqrt{y} \) involves the square root of \( y \). For \( y < 0 \), the square root function is not defined for real numbers, as square roots of negative numbers are imaginary. This means that if the initial condition \( y_0 < 0 \), there will be no real-valued function \( y(x) \) that satisfies the given differential equation.

Step 2: Solve the Differential Equation for y_0 > 0

Given the differential equation \( \frac{dy}{dx} = \sqrt{y} \), we can separate variables to solve it. We rewrite it as \( \frac{1}{\sqrt{y}} dy = dx \). Integrating both sides, we have:\[ \int \frac{1}{\sqrt{y}} dy = \int dx \]This results in:\[ 2\sqrt{y} = x + C \] where \( C \) is the constant of integration.

Step 3: Applying the Initial Condition to Find the Constant

Use the initial condition \( y(x_0) = y_0 \) to solve for \( C \):\[ 2\sqrt{y_0} = x_0 + C \]Solving for \( C \), we find:\[ C = 2\sqrt{y_0} - x_0 \]

Step 4: Expressing y in Terms of x

Substituting \( C \) back into the expression, we have:\[ 2\sqrt{y} = x - x_0 + 2\sqrt{y_0} \]Solving for \( y \), we square both sides:\[ y = \left( \frac{x - x_0 + 2\sqrt{y_0}}{2} \right)^2 \]

Step 5: Determine the Largest Interval for the Solution

The solution is defined only when the square root is non-negative, hence:\[ \frac{x - x_0 + 2\sqrt{y_0}}{2} \geq 0 \]Solving this inequality,\[ x \geq x_0 - 2\sqrt{y_0} \]Therefore, the solution is valid for \( x \) in the interval \( [x_0 - 2\sqrt{y_0}, \infty) \).

Key Concepts

Initial Value Problem

An initial value problem (IVP) in the context of ordinary differential equations involves finding a function that satisfies a differential equation and an initial condition. The initial condition is typically specified as the value of the function at a particular point. For instance, in our problem, we have \( \frac{d y}{d x}=\sqrt{y} \) with the condition \( y(x_0) = y_0 \). This means we need to find a function \( y(x) \) that not only satisfies the differential equation but also equals \( y_0 \) when \( x = x_0 \). If the initial condition falls outside the domain where the differential equation is defined, the IVP may have no real solutions, as is the case when \( y_0 < 0 \) here. Thus, properly understanding and applying initial conditions is crucial for solving IVPs.

Separation of Variables

Separation of variables is a method used to solve differential equations by separating the variables into two sides of the equation. This approach is particularly useful when dealing with ordinary differential equations like \( \frac{dy}{dx} = \sqrt{y} \). Here's how it works: you rewrite the equation so that each side contains only one variable. In this case, by rearranging, we achieve \( \frac{1}{\sqrt{y}} dy = dx \). This allows us to integrate each side separately:

• Integrate \( \int \frac{1}{\sqrt{y}} dy \) to get \( 2\sqrt{y} \).
• Integrate \( \int dx \) to get \( x + C \).

After integration, solving for \( y \) in terms of \( x \) gives us the solution that meets the initial condition.

Existence of Solutions

The existence of solutions for an initial value problem is fundamental. Not every differential equation with initial conditions has a solution, particularly under certain conditions concerning the initial values themselves. For the equation \( \frac{dy}{dx} = \sqrt{y} \), solutions exist only when \( y_0 \geq 0 \). The reason is that the function \( \sqrt{y} \) is only defined for non-negative \( y \), as negative values would result in imaginary numbers. Thus, acknowledging the domain of the function and the influence of initial conditions is imperative. The statement "no solution for \( y_0 < 0 \)" hinges on the non-existence of square roots of negative numbers in the set of real numbers.

Interval of Validity

When solving an ordinary differential equation, determining the interval of validity is essential. This interval describes where the solution is reliably defined. For the initial value problem given by \( \frac{dy}{dx} = \sqrt{y} \), solving for \( y \) yields solutions, assuming \( y_0 > 0 \). After applying the initial condition, we find that the interval of validity depends on ensuring that under the root, values remain non-negative:

• The expression \( \frac{x - x_0 + 2\sqrt{y_0}}{2} \geq 0 \) leads to the interval \( [x_0 - 2\sqrt{y_0}, \infty) \).
• This means the solution remains valid from the point \( x_0 - 2\sqrt{y_0} \) and onwards.

The solution's interval of validity contains all \( x \) values that keep the expression within a real and solvable domain for \( y(x) \).