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Verify the following general solutions and find the particular solution. Find the particular solution to the differential equation \(y^{\prime}=2 x y\) that passes through \(\left(0, \frac{1}{2}\right)\), given that \(y=C e^{x^{2}}\) is a general solution.

Short Answer

Expert verified
The particular solution is \( y = \frac{1}{2} e^{x^2} \).

Step by step solution

01

Understand the General Solution

Given the general solution to the differential equation as \( y = C e^{x^2} \), we need to first verify that it solves the differential equation \( y' = 2xy \).
02

Differentiate the General Solution

Differentiate \( y = C e^{x^2} \) with respect to \( x \) to get \( y' = \frac{d}{dx}[C e^{x^2}] = C \cdot \frac{d}{dx}[e^{x^2}] = C \cdot 2x e^{x^2} = 2x(C e^{x^2}) = 2xy \), which matches the given differential equation \( y' = 2xy \).
03

Substitute Initial Conditions

Use the initial condition \( y = \frac{1}{2} \) when \( x = 0 \). Substitute into the general solution: \( \frac{1}{2} = C e^{0^2} \). Simplify to find \( C = \frac{1}{2} \).
04

Write the Particular Solution

Substitute the value of \( C \) back into the general solution, obtaining \( y = \frac{1}{2} e^{x^2} \), which is the particular solution.

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Key Concepts

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

Particular Solution
In the realm of differential equations, a particular solution is a specific solution that fits the unique conditions of a problem, often called initial conditions. It means that while the differential equation might have many solutions, by applying initial conditions, we find one precise solution. This is crucial for real-world applications where the specific scenario is known.

The particular solution comes from the general solution, which encompasses a family of solutions through a parameter, usually denoted as \( C \). Here, for the differential equation \( y^{\prime}=2 x y \) and the general solution \( y=C e^{x^{2}} \), we find the particular version by substituting initial conditions. Thus, by applying the initial condition that the solution passes through \( (0, \frac{1}{2}) \), we determine \( C = \frac{1}{2} \).
  • Particular solution: \( y = \frac{1}{2} e^{x^2} \)
It narrows down the wide array of possibilities from the general solution to just one, providing a perfect fit for the specified conditions.
Initial Conditions
Initial conditions are like the secret key that unlocks a specific scenario for a differential equation. They define the starting point or particular case for which we need to find specific solutions. When a differential equation is solved, there are generally an infinite number of potential solutions. With the initial conditions, we have additional information that allows us to narrow it down to just one.

In our example, the initial condition was that the solution passes through the point \((0, \frac{1}{2})\). Therefore, when \( x = 0 \), \( y = \frac{1}{2} \). This condition enabled us to calculate the specific value of \( C \) in the general solution \( y = C e^{x^2} \).
  • Substitute: \( y = \frac{1}{2}, x = 0 \)
  • Equation becomes: \( \frac{1}{2} = C \cdot e^{0} \)
  • Find \( C = \frac{1}{2} \)
Initial conditions help transform a general blueprint into a personalized, precise prediction for the given situation.
General Solution
The general solution of a differential equation is like a family of solutions, spanning an entire spectrum through an arbitrary constant, often represented by \( C \). This constant accommodates various solutions from which the particular solution can be derived once additional information such as initial conditions is provided.

For the differential equation \( y^{\prime}=2 x y \), the general solution was given as \( y = C e^{x^{2}} \). This represents all possible solutions without specific conditions applied, allowing the solution to match different scenarios by choosing different values for \( C \).
  • General form: \( y = C e^{x^{2}} \)
By applying initial conditions, this general formula is tailored to the needs of a specific problem, thus highlighting the broad applicability of differential equations to various contexts and scenarios.

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Most popular questions from this chapter

[T] The population of trout in a pond is given by \(P^{\prime}=0.4 P\left(1-\frac{P}{10000}\right)-400, \quad\) where 400 trout are caught per year. Use your calculator or computer software to draw a directional field and draw a few sample solutions. What do you expect for the behavior?

For the following problems, use a software program or your calculator to generate the directional fields. Solve explicitly and draw solution curves for several initial conditions. Are there some critical initial conditions that change the behavior of the solution? $$ [\mathbf{T}] y^{\prime}=y^{2} x^{3} $$

Torricelli's law states that for a water tank with a hole in the bottom that has a cross-section of \(A\) and with a height of water \(h\) above the bottom of the tank, the rate of change of volume of water flowing from the tank is proportional to the square root of the height of water, according to \(\frac{d V}{d t}=-A \sqrt{2 g h},\) where \(g\) is the acceleration due to gravity. Note that \(\frac{d V}{d t}=A \frac{d h}{d t} .\) Solve the resulting initial-value problem for the height of the water, assuming a tank with a hole of radius \(2 \mathrm{ft}\). The initial height of water is \(100 \mathrm{ft}\).

For the following problems, find the solution to the initial value problem. $$ x y^{\prime}=y(x-2), y(1)=3 $$

A falling object of mass \(m\) can reach terminal velocity when the drag force is proportional to its velocity, with proportionality constant \(k\). Set up the differential equation and solve for the velocity given an initial velocity of 0

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