Chapter 8: Problem 16

By separation of variables, find a solution of the equation \(y^{\prime}=\sqrt{y}\) containing one arbitrary constant. Find a particular solution satisfying \(y=0\) when \(x=0 .\) Show that \(y \equiv 0\) is a solution of the differential equation which cannot be obtained by specializing the arbitrary constant in your solution above. Computer plot a slope field and some of the solution curves. Show that there are an infinite number of solution curves passing through any point on the \(x\) axis, but just one through any point for which \(y>0 .\) Hint: See Example 3. Problems 17 and 18 are physical problems leading to this differential equation.

Short Answer

Expert verified

General solution: \( y = \left( \frac{x + C}{2} \right)^2 \). Particular solution: \( y = \frac{x^2}{4} \), when \( y(0) = 0 \). Solution \( y \equiv 0 \) is also valid but not obtainable by setting \(C\) in the general solution.

Step by step solution

Separate Variables

Start by separating the variables in the differential equation. Given: \( y' = \sqrt{y} \).Rewrite it as \( \frac{dy}{dx} = \sqrt{y} \). Separate variables to get \( \frac{dy}{\sqrt{y}} = dx \).

Integrate Both Sides

Integrate both sides of the equation. For the left side, perform the following integration: \( \int \frac{1}{\sqrt{y}} dy = \int dx \). Rewrite the left side as: \(\int y^{-1/2} dy \).

Apply Integration

Apply the integration on both sides:\[ \int y^{-1/2} dy = 2 y^{1/2} + C_1 \] and \( \int dx = x + C_2 \).

Combine Constants

Combine the constants into a single constant, say, \(C\) and set the equal results of the integrations:\( 2 y^{1/2} = x + C \).

Solve for y

Solve for \( y \):\[ y = \left( \frac{x + C}{2} \right)^2. \]

Find Particular Solution

Given the initial condition \( y = 0 \) when \( x = 0 \), substitute these into the equation to determine the particular value for \(C\): \[ 0 = \left( \frac{0 + C}{2} \right)^2 \Rightarrow C = 0 \]. Thus, the particular solution becomes \( y = \left( \frac{x}{2} \right)^2 \Rightarrow y = \frac{x^2}{4} \).

Verify y ≡ 0 as a Solution

Show that \( y \equiv 0 \) is a solution of the differential equation by substituting back into the original equation: \( y' = \sqrt{y} \rightarrow 0 = \sqrt{0} \Rightarrow 0 = 0 \). Hence, \( y \equiv 0 \) is a solution.

Plot Slope Field and Solution Curves

Plot a slope field using the differential equation \( y' = \sqrt{y} \). Draw some of the solution curves for different initial conditions. Observe that there are infinite solution curves passing through any point on the \(x\)-axis but only one solution curve for each point where \( y > 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Separation of Variables

Separation of variables is a key technique used to solve differential equations. We begin by isolating the variables on different sides of the equation. In the given equation, we have:

Start with the equation: \( y' = \sqrt{y} \).
Rewrite it as \( \frac{dy}{dx} = \sqrt{y} \).
Separate the terms to get: \( \frac{dy}{\sqrt{y}} = dx \).

By doing this, we make it possible to integrate each side independently, a step we’ll discuss next.

Integration

Integration is the process of finding the antiderivative of a function. After separating the variables, the next step is to integrate both sides of the equation:

For the left side: \( \int \frac{1}{\sqrt{y}} dy \).
Rewrite it as: \( \int y^{-1/2} dy \).
For the right side, integrate \( dx \).

Performing these integrals, we get:

\( 2 y^{1/2} + C_1 \) from the left side.
\( x + C_2 \) from the right side.

Combine these results into a single equation with a constant \( C \): \( 2 y^{1/2} = x + C \). Finally, to solve for \( y \), square both sides: \( y = \left(\frac{x + C}{2} \right)^2 \). Now, we have a general solution containing an arbitrary constant.

Initial Conditions

Initial conditions are specific values for which a solution to a differential equation needs to be found. These conditions help us determine the constant in our general solution. Given: \( y = 0 \) when \( x = 0 \), substitute these into the equation:

\( 0 = \left(\frac{0 + C}{2} \right)^2 \).
Solve for \( C \): \( C = 0 \).

Thus, the particular solution is: \( y = \left(\frac{x}{2}\right)^2 \Rightarrow y = \frac{x^2}{4} \). Initial conditions are essential in narrowing down from the general solution to a specific one.

Slope Field and Solution Curves

A slope field is a graphical representation of the slopes of a differential equation at given points. It helps visualize the behavior of solutions without solving the equation analytically. To plot the slope field for our differential equation \( y' = \sqrt{y} \):

Draw small line segments with slopes equal to \( \sqrt{y} \) at various points in the coordinate plane.
Observe the pattern these slopes create.

From our observations:

There are infinite solution curves passing through any point on the \( x \)-axis (where \( y=0 \)).
There is only one unique solution curve passing through any point where \( y > 0 \).

Plotting these curves helps us understand the family of solutions for the differential equation.

Conclusion

By using separation of variables and integration, we derived the general solution \( y = \left(\frac{x + C}{2} \right)^2 \). Applying the initial condition \( y = 0 \) when \( x = 0 \), we found the particular solution \( y = \frac{x^2}{4} \). We also verified that \( y \equiv 0 \) is indeed a solution. Finally, plotting the slope field and solution curves provided us with a visual representation of the solutions, showing an infinite number of curves passing through points on the \( x \)-axis but only one through points where \( y > 0 \).