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The solution of dy/dx = y2 with initial value y(0) = 1 is bounded in the interval
- a)−∞ ≤ x ≤ ∞
- b)−∞ ≤ x ≤ 1
- c)x < 1, x > 1
- d)−2 ≤ x ≤ 2
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
The solution of dy/dx = y2with initial value y(0) = 1 is bounded in th...
⇒ from the Integrating the equation, we get
⇒ at x = 0, y = 1, hence -1/1 = c ⇒ c = -1
x ≠ 1, at (x = 1) y will be undefined
x > 1, x < 1
Most Upvoted Answer
The solution of dy/dx = y2with initial value y(0) = 1 is bounded in th...
Understanding the Differential Equation
The given differential equation is:
\[
\frac{dy}{dx} = y^2
\]
This is a separable differential equation, which we can solve by separating variables.
Solving the Differential Equation
1. **Separate the Variables**:
\[
\frac{1}{y^2} dy = dx
\]
2. **Integrate Both Sides**:
- Left Side: \(-\frac{1}{y} + C_1\)
- Right Side: \(x + C_2\)
Combining these gives:
\[
-\frac{1}{y} = x + C
\]
3. **Rearranging**:
\[
y = -\frac{1}{x + C}
\]
4. **Applying the Initial Condition \(y(0) = 1\)**:
- Plugging in \(x = 0\):
\[
1 = -\frac{1}{0 + C} \implies C = -1
\]
Thus, the solution is:
\[
y = -\frac{1}{x - 1}
\]
Behavior of the Solution
- The solution \(y = -\frac{1}{x - 1}\) is undefined at \(x = 1\) (vertical asymptote).
- For \(x < 1\),="" \(y\)="" is="" negative="" and="" decreases="" without="" bound="" as="" \(x\)="" approaches="" 1="" from="" the="" />
- For \(x > 1\), \(y\) is negative and approaches 0 as \(x\) increases.
Conclusion on Boundedness
- The solution is bounded in the interval \(x < 1\)="" and="" \(x="" /> 1\), but not at \(x = 1\) where it is undefined.
Thus, the correct option is **C**: \(x < 1,="" x="" /> 1\).
The given differential equation is:
\[
\frac{dy}{dx} = y^2
\]
This is a separable differential equation, which we can solve by separating variables.
Solving the Differential Equation
1. **Separate the Variables**:
\[
\frac{1}{y^2} dy = dx
\]
2. **Integrate Both Sides**:
- Left Side: \(-\frac{1}{y} + C_1\)
- Right Side: \(x + C_2\)
Combining these gives:
\[
-\frac{1}{y} = x + C
\]
3. **Rearranging**:
\[
y = -\frac{1}{x + C}
\]
4. **Applying the Initial Condition \(y(0) = 1\)**:
- Plugging in \(x = 0\):
\[
1 = -\frac{1}{0 + C} \implies C = -1
\]
Thus, the solution is:
\[
y = -\frac{1}{x - 1}
\]
Behavior of the Solution
- The solution \(y = -\frac{1}{x - 1}\) is undefined at \(x = 1\) (vertical asymptote).
- For \(x < 1\),="" \(y\)="" is="" negative="" and="" decreases="" without="" bound="" as="" \(x\)="" approaches="" 1="" from="" the="" />
- For \(x > 1\), \(y\) is negative and approaches 0 as \(x\) increases.
Conclusion on Boundedness
- The solution is bounded in the interval \(x < 1\)="" and="" \(x="" /> 1\), but not at \(x = 1\) where it is undefined.
Thus, the correct option is **C**: \(x < 1,="" x="" /> 1\).
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Community Answer
The solution of dy/dx = y2with initial value y(0) = 1 is bounded in th...
⇒ from the Integrating the equation, we get
⇒ at x = 0, y = 1, hence -1/1 = c ⇒ c = -1
x ≠ 1, at (x = 1) y will be undefined
x > 1, x < 1
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The solution of dy/dx = y2with initial value y(0) = 1 is bounded in the intervala)−∞ ≤ x ≤ ∞b)−∞ ≤ x ≤ 1c)x < 1,x > 1d)−2 ≤ x ≤ 2Correct answer is option 'C'. Can you explain this answer?
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