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In Exercises 7 and 8, the general solution of a differential equation is given. (a) Find the particular solution that satisfies the given initial condition. (b) Plot the solution curves correspond. ing to the given values of \(C\). Indicate the solution curve that corresponds to the solution found in part (a). $$ \begin{array}{l} y \frac{d y}{d x}-e^{2 x}=0, \quad y^{2}=e^{2 x}+C ; \quad y(0)=1 ; \\ C=-2,-1,0,1,2 \end{array} $$

Short Answer

Expert verified
The particular solution that satisfies the given initial condition \(y(0) = 1\) is: \(y^2 = e^{2x}\). When plotting the solution curves for the given values of \(C\), the curves shift upward as the value of \(C\) increases. The particular solution found in Step 1 corresponds to the curve where \(C = 0\).

Step by step solution

01

Find the particular solution using the initial condition

We are given the general solution: \(y^2 = e^{2x} + C\) and the initial condition: \(y(0) = 1\). We need to find the value of \(C\) for which this initial condition holds true. Plug in the initial condition into the general solution: \(1^2 = e^{2 \cdot 0} + C\) Solve for \(C\) : \(1 = 1 + C\) \(C = 1 - 1 = 0\) So, the particular solution that satisfies the given initial condition \(y(0) = 1\) is: \(y^2 = e^{2x}\)
02

Plot the solution curves for the given values of \(C\)

We will now plot the solution curves corresponding to the given values of \(C\): -2, -1, 0, 1, 2. 1. For \(C=-2\), the solution is \(y^2 = e^{2x} -2\) 2. For \(C=-1\), the solution is \(y^2 = e^{2x} -1\) 3. For \(C=0\), the solution is \(y^2 = e^{2x}\) (This is the solution curve for the particular solution found in step 1). 4. For \(C=1\), the solution is \(y^2 = e^{2x} + 1\) 5. For \(C=2\), the solution is \(y^2 = e^{2x} + 2\) Plotting these solutions, we can see that the curves shift upward as the value of \(C\) increases. Keep in mind that the curve for the particular solution will be \(y^2 = e^{2x}\) (where \(C\) = 0).

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Key Concepts

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

Initial Condition
The initial condition of a differential equation provides a starting point from which we can determine a specific, or particular solution. This initial condition generally consists of a specific value for the independent variable and the corresponding value of the solution (dependent variable). Understanding the initial condition is crucial since different initial conditions can lead to entirely different behaviors and paths of a solution.

For instance, in our exercise, the initial condition is given as \( y(0) = 1 \). When we have the general solution \( y^2 = e^{2x} + C \), this condition helps us find the exact value of \( C \) that corresponds to our particular case. Substituting \((x, y)\) coordinates from the initial condition into the general solution, we are able to isolate and solve for the constant \( C \), which leads us to the particular solution that matches the condition provided at the beginning.
Particular Solution
A particular solution of a differential equation is a solution that satisfies both the differential equation and an initial condition. Each particular solution can be seen as a single path that the function takes based on a specific starting point. In contrast to the general solution, which represents an entire family of curves defined by a constant \( C \), the particular solution is unique to its initial condition.

Referring back to our example, after determining the constant \( C \) as 0, we plugged it back into the general formula resulting in \( y^2 = e^{2x} \), the particular solution for the initial condition \( y(0) = 1 \). This is a specific instance of the broader family of solutions to the differential equation, giving a defined trajectory for the values of \( y \) as \( x \) varies.
Plotting Solution Curves
Plotting solution curves for a differential equation provides a visual representation of how solutions behave for varying values of \( C \). Each curve corresponds to a unique solution, and together, they form a family of curves known as solution curves.

In the exercise, we were directed to plot solution curves corresponding to different values of the constant \( C \). Here's a step-by-step guide for the visualization process:
  • Determine the range of x values you want to plot.
  • Calculate the corresponding values of \( y \) for each \( x \), for every value of \( C \).
  • For each value of \( C \), plot the points on the same graph to see how they relate to one another.
  • Highlight the particular curve that matches the initial condition. In our case, this would be the curve where \( C = 0 \).

Through plotting, we can observe that the constant \( C \) vertically shifts the curve on the graph. The graphical representation helps to contextualize the abstract mathematical concepts and deepens comprehension by illustrating the influence of different constants on the shape and location of the solution curves.

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Most popular questions from this chapter

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