Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Find the "general solution" of the equation \(y^{\prime}=\sqrt{y}\) by separation of variables. Find a particular solution satisfying \(y=0\) when \(x=0\). Show that the singular solution \(y=0\) cannot be obtained from the general solution. Sketch graphs of the "general solution" for several values of the arbitrary constant, and observe that each of them is tangent to the singular solution. Thus there are two solutions passing through any point on the \(x\) axis; in particular, there are two solutions satisfying \(x=y=0 .\) Problems 17 and 18 are physical problems leading to this differential equation.

Short Answer

Expert verified
General solution is \( y = \frac{(x + C)^2}{4} \). The particular solution for \( y(0) = 0 \) is \( y = \frac{x^2}{4} \). Singular solution \( y = 0 \) cannot be obtained from the general solution.

Step by step solution

01

Write the Differential Equation

Start with the given differential equation: \[ y' = \sqrt{y} \] We will solve this using separation of variables.
02

Separate the Variables

Separate the variables such that all terms involving \(y\) are on one side and all terms involving \( x \) are on the other side. \[ \frac{dy}{\sqrt{y}} = dx \]
03

Integrate Both Sides

Integrate both sides of the equation: \[ \int \frac{1}{\sqrt{y}} dy = \int dx \] This results in: \[ 2 \sqrt{y} = x + C \] Where \( C \) is the integration constant.
04

Solve for y

Solve for \( y \) to find the general solution: \[ \sqrt{y} = \frac{x + C}{2} \] Square both sides: \[ y = \left( \frac{x + C}{2} \right)^2 \] Simplifying, we get: \[ y = \frac{(x + C)^2}{4} \]
05

Find Particular Solution for Given Conditions

To find a particular solution satisfying \( y(0) = 0 \), plug \( x = 0 \) and \( y = 0 \) into the general solution equation: \[ 0 = \frac{(0 + C)^2}{4} \] This simplifies to \[ C = 0 \]. Therefore, the particular solution is: \[ y = \frac{x^2}{4} \]
06

Confirm the Singular Solution

The singular solution is \( y = 0 \). Notice that for \( y = 0 \), the right-hand side of the original differential equation \( y' = \sqrt{y} \) is also 0. However, \( y = 0 \) cannot be derived from the general solution \( y = \frac{(x + C)^2}{4} \) because for any value of \( C \), the general solution does not satisfy \( y = 0 \) for any value of \( x \). Therefore, \( y = 0 \) is a unique solution not represented by the general solution.
07

Sketch the Graphs

Plot the general solutions \( y = \frac{(x + C)^2}{4} \) for several values of \( C \). Observe that each curve is parabolic in shape and is tangent to the singular solution \( y = 0 \) at exactly one point.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

separation of variables
Separation of variables is a straightforward technique used to solve first-order differential equations where variables can be rearranged. Take the differential equation given in our problem: \( y' = \sqrt{y} \). The idea is to move all terms involving \( y \) to one side and all terms involving \( x \) to the other side:
\[ \frac{dy}{\sqrt{y}} = dx \] Now, each side of the equation can be integrated independently. This process allows us to solve the equation step by step.

It's like untangling two intertwined strings by moving them apart, making integration simpler.
particular solution
A particular solution is a specific solution derived from the general solution by applying initial conditions or boundary conditions. For example, given the condition \( y(0) = 0 \), substitute into the general solution to find a specific value for the constant:
\[ 0 = \frac{\left( 0 + C \right)^2}{4} \] Simplifying this gives us \( C = 0 \). Thus, the particular solution becomes:
\[ y = \frac{x^2}{4} \]
The particular solution uniquely fits the given initial conditions, providing a specific trajectory for the equation.
singular solution
A singular solution is a solution to a differential equation that is not part of the general solution family but still satisfies the equation. In our example, \( y = 0 \) is a singular solution.
It’s special because it cannot be obtained by any value of the integration constant \( C \) from the general solution \( y = \frac{(x + C)^2}{4} \). Singular solutions often reveal unique behaviors in differential equations, undetected by general solutions.
They might represent boundary behaviors or other special cases, providing deeper insights into the system's dynamics. Models can sometimes behave unexpectedly at singular points, so identifying these solutions is crucial.
integration constant
The integration constant \( C \) appears when we integrate an indefinite integral. It represents an entire family of solutions rather than just one specific solution.
In solving \( y' = \sqrt{y} \), we integrated both sides to get:
\[ 2 \sqrt{y} = x + C \]
This constant \( C \) signifies that there are infinitely many solutions based on different initial conditions. Each specific value of \( C \) corresponds to a different curve or trajectory in the \( y \)-\( x \)-plane.
graphical representation
Graphical representation involves plotting the solutions on a graph to visualize their behavior. For our general solution \( y = \frac{(x + C)^2}{4} \), each specific value of \( C \) will define a parabolic curve. Plotting these curves for several values of \( C \), you will notice:
  • Each curve is parabolic.
  • Each curve is tangent to the singular solution \( y = 0 \) at exactly one point.

This tangency is significant because it shows that while the curves approach the zero solution, they never fully align, emphasizing the uniqueness of the singular solution. Visualizing differential equations can often give better intuition of the solution behaviors.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solve the following differential equations. . \(\left(D^{2}-2 D+1\right) y=0\)

The momentum \(p\) of an electron at velocity \(v\) near the velocity \(c\) of light increases according to the formula $$ p=\frac{m_{0} v}{\sqrt{1-v^{2} / c^{2}}} $$ where \(m_{0}\) is a constant (the rest mass). If an electron is subject to a constant force \(F\), Newton's second law describing its motion is $$ \frac{d p}{d t}=\frac{d}{d t}\left(\frac{m_{0} v}{\sqrt{1-v^{2} / c^{2}}}\right)=F $$ Find the velocity as a function of time and show that the limiting velocity as \(t\) tends to infinity is \(c\). Find the distance traveled by the electron in time \(t\) if it starts from rest.

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \((2 x-y \sin 2 x) d x+\left(2 y-\sin ^{2} x\right) d y=0\)

Solve the following differential equations. \(y^{\prime}+y=x y^{2 / 3}\)

An object of mass \(m\) falls from rest under gravity subject to an air resistance proportional to its velocity. Taking the \(y\) axis as positive down, show that the differential equation of motion is \(m(d v / d t)=m g-k v\), where \(k\) is a positive constant. Find \(v\) as a function of \(t\), and find the limiting value of \(v\) as \(t\) tends to infinity; this limit is called the terminal relocity. Can you find the terminal velocity directly from the differential equation without solving it? Hint: What is \(d v / d t\) after \(v\) has reached an essentially constant value? Consider the following specific examples of this problem. (a) A person drops from an airplane with a parachute. Find a reasonable value of \(k\). (b) In the Millikan oil drop experiment to measure the charge of an electron, tiny electrically charged drops of oil fall through air under gravity or rise under the combination of gravity and an electric field. Measurements can be made only after they have reached terminal velocity. Find a formula for the time required for a drop starting at rest to reach \(99 \%\) of its terminal velocity.

See all solutions

Recommended explanations on Combined Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free