Problem 38
Question
(a) The second derivative of \(y\) is \\[\frac{d^{2} y}{d x^{2}}=-\frac{d y / d x}{(y-1)^{2}}=-\frac{1 /(y-3)}{(y-3)^{2}}=-\frac{1}{(y-3)^{3}}\\] The solution curve is concave down when \(d^{2} y / d x^{2}<0\) or \(y>3\), and concave up when \(d^{2} y / d x^{2}>0\) or \(y<3 .\) From the phase portrait we see that the solution curve is decreasing when \(y<3\) and increasing when \(y>3\). (b) Separating variables and integrating we obtain \\[ \begin{aligned} (y-3) d y &=d x \\ \frac{1}{2} y^{2}-3 y &=x+c \\ y^{2}-6 y+9 &=2 x+c_{1} \\ (y-3)^{2} &=2 x+c_{1} \\ y &=3 \pm \sqrt{2 x+c_{1}}.\end{aligned}\\] The initial condition dictates whether to use the plus or minus sign. When \(y_{1}(0)=4\) we have \(c_{1}=1\) and \(y_{1}(x)=3+\sqrt{2 x+1}\). When \(y_{2}(0)=2\) we have \(c_{1}=1\) and \(y_{2}(x)=3-\sqrt{2 x+1}.\) When \(y_{3}(1)=2\) we have \(c_{1}=-1\) and \(y_{3}(x)=3-\sqrt{2 x-1}.\) When \(y_{4}(-1)=4\) we have \(c_{1}=3\) and \(y_{4}(x)=3+\sqrt{2 x+3}\).
Step-by-Step Solution
Verified Answer
The concavity and specific solution for each initial condition depend on the value of \(y\) in relation to 3 and given \(c_1\).
1Step 1: Analyze the Second Derivative
Given the second derivative \(\frac{d^2 y}{d x^2} = -\frac{1}{(y-3)^3}\), determine the concavity based on its sign. The curve is concave down when \(\frac{d^2 y}{d x^2} < 0\), occurring when \(y > 3\). The curve is concave up when \(\frac{d^2 y}{d x^2} > 0\), happening when \(y < 3\).
2Step 2: Determine the Behavior from Phase Portrait
The solution curve decreases when \(y < 3\) and increases when \(y > 3\) as seen from the phase portrait.
3Step 3: Separate Variables and Integrate
Separate variables and integrate: \((y-3)\, dy = dx\). Integrating gives: \(\frac{1}{2} y^2 - 3y = x + c\), which simplifies to \((y-3)^2 = 2x + c_1\).
4Step 4: Solve for \(y\)
Solving \((y-3)^2 = 2x + c_1\), we obtain \(y = 3 \pm \sqrt{2x + c_1}\). The choice of plus or minus depends on initial conditions.
5Step 5: Apply Initial Conditions
Using the given initial conditions: - For \(y_1(0) = 4\), \(c_1 = 1\), resulting in \(y_1(x) = 3 + \sqrt{2x + 1}\).- For \(y_2(0) = 2\), \(c_1 = 1\), yielding \(y_2(x) = 3 - \sqrt{2x + 1}\).- For \(y_3(1) = 2\), \(c_1 = -1\), leading to \(y_3(x) = 3 - \sqrt{2x - 1}\).- For \(y_4(-1) = 4\), \(c_1 = 3\), resulting in \(y_4(x) = 3 + \sqrt{2x + 3}\).
Key Concepts
Concavity AnalysisPhase PortraitSeparation of VariablesInitial Conditions
Concavity Analysis
In the realm of differential equations, understanding the concavity of function curves is crucial. Concavity helps us determine how the curve bends at every point. By looking at the second derivative, \(\frac{d^2 y}{d x^2}\), we can infer the concavity of the curve:
- If \(\frac{d^2 y}{d x^2} < 0\), the curve is concave down, meaning it curves like a frown. In this particular problem, this condition is fulfilled when \(y > 3\).
- Conversely, if \(\frac{d^2 y}{d x^2} > 0\), the curve is concave up, resembling a smile, happening when \(y < 3\).
This understanding helps us predict how the solution curve behaves, aiding in sketching accurate graphs.
- If \(\frac{d^2 y}{d x^2} < 0\), the curve is concave down, meaning it curves like a frown. In this particular problem, this condition is fulfilled when \(y > 3\).
- Conversely, if \(\frac{d^2 y}{d x^2} > 0\), the curve is concave up, resembling a smile, happening when \(y < 3\).
This understanding helps us predict how the solution curve behaves, aiding in sketching accurate graphs.
Phase Portrait
A phase portrait offers a powerful visual tool for analyzing differential equations. By graphically representing a system's trajectories, phase portraits deliver insights into the solution's overall behavior.
In our exercise:
- The phase portrait indicates that the solution curve decreases when \(y < 3\).
- Meanwhile, it shows an increase for \(y > 3\).
These visual cues are invaluable. They not only support algebraic findings but also enhance intuition about how solutions evolve over time.
In our exercise:
- The phase portrait indicates that the solution curve decreases when \(y < 3\).
- Meanwhile, it shows an increase for \(y > 3\).
These visual cues are invaluable. They not only support algebraic findings but also enhance intuition about how solutions evolve over time.
Separation of Variables
Separation of variables is a fundamental technique utilized to solve differential equations. This approach involves separating the equation's variables to integrate each side independently.
In the context of this problem, we start by writing:
\[(y-3)dy = dx\]
The next steps involve integrating both sides to reveal a relationship between\(x\) and \(y\):
- Integrating results in \(\frac{1}{2}y^2 - 3y = x + c\), which simplifies to \((y-3)^2 = 2x + c_1\).
This solves the equation implicitly for \(y\), reducing the complexity of solving more challenging differential equations.
In the context of this problem, we start by writing:
\[(y-3)dy = dx\]
The next steps involve integrating both sides to reveal a relationship between\(x\) and \(y\):
- Integrating results in \(\frac{1}{2}y^2 - 3y = x + c\), which simplifies to \((y-3)^2 = 2x + c_1\).
This solves the equation implicitly for \(y\), reducing the complexity of solving more challenging differential equations.
Initial Conditions
Initial conditions are established values that allow us to find a specific solution from a general family of solutions. Without these, the solution would remain ambiguous due to the arbitrary constant introduced during integration.
In our example:
- When given \(y_1(0) = 4\), we find \(c_1 = 1\) resulting in \(y_1(x) = 3+\sqrt{2x+1}\).
- For \(y_2(0) = 2\), \(c_1 = 1\), leading to \(y_2(x) = 3-\sqrt{2x+1}\).
- Given \(y_3(1) = 2\), \(c_1 = -1\) and \(y_3(x) = 3-\sqrt{2x-1}\).
- Lastly, \(y_4(-1)=4\) gives \(c_1 = 3\) and \(y_4(x) = 3 + \sqrt{2x+3}\).
These specific solutions depict the importance of initial conditions in defining the behavior of a differential system fully.
In our example:
- When given \(y_1(0) = 4\), we find \(c_1 = 1\) resulting in \(y_1(x) = 3+\sqrt{2x+1}\).
- For \(y_2(0) = 2\), \(c_1 = 1\), leading to \(y_2(x) = 3-\sqrt{2x+1}\).
- Given \(y_3(1) = 2\), \(c_1 = -1\) and \(y_3(x) = 3-\sqrt{2x-1}\).
- Lastly, \(y_4(-1)=4\) gives \(c_1 = 3\) and \(y_4(x) = 3 + \sqrt{2x+3}\).
These specific solutions depict the importance of initial conditions in defining the behavior of a differential system fully.
Other exercises in this chapter
Problem 38
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