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What is the reason behind the non-existence of any real function which satisfies the differential equation, (y’)2 + 1 = 0?
- a)Because for any real function, the left-hand side of the equation will be greater than, or equal to one and thus cannot be zero
- b)Because for any real function, the left-hand side of the equation will be less than, or equal to one and thus cannot be zero
- c)Because for any real function, the left-hand side of the equation becomes zero
- d)Because for any real function, the left-hand side of the equation becomes infinity
Correct answer is option 'A'. Can you explain this answer?
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What is the reason behind the non-existence of any real function which...
Given: (y’)2 + 1 = 0 Consider if y = 2x, then y’ = 2 and hence the left-hand side of the equation becomes 3 which is greater than 1. Therefore, the left-hand side of the equation will always be greater than, or equal to one and thus cannot be zero and hence the differential equation is not satisfied.
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What is the reason behind the non-existence of any real function which...
There could be several reasons why a real function may not exist that satisfies a given differential equation. Here are a few possibilities:
1. Inconsistent initial conditions: The differential equation may have certain initial conditions that are not consistent with the existence of a real solution. For example, if the initial condition requires the value of the function to be complex, then a real solution may not exist.
2. Singularities: The differential equation may have singularities or points where the function is not defined. These singularities can prevent the existence of a real solution. For example, if a differential equation has a singularity at a certain point, then a real solution may not exist in the vicinity of that point.
3. Incompatible constraints: The differential equation may have constraints or conditions that are not compatible with the existence of a real solution. For example, if a differential equation requires the function to be periodic with a certain period that is not compatible with real numbers, then a real solution may not exist.
4. Nonlinearity: The differential equation may be nonlinear, meaning that it involves higher order terms or products of the function and its derivatives. Nonlinear differential equations can be more difficult to solve and may not always have real solutions.
These are just a few possible reasons, and the specific reason for the non-existence of a real solution would depend on the specific form of the differential equation and its associated conditions.
1. Inconsistent initial conditions: The differential equation may have certain initial conditions that are not consistent with the existence of a real solution. For example, if the initial condition requires the value of the function to be complex, then a real solution may not exist.
2. Singularities: The differential equation may have singularities or points where the function is not defined. These singularities can prevent the existence of a real solution. For example, if a differential equation has a singularity at a certain point, then a real solution may not exist in the vicinity of that point.
3. Incompatible constraints: The differential equation may have constraints or conditions that are not compatible with the existence of a real solution. For example, if a differential equation requires the function to be periodic with a certain period that is not compatible with real numbers, then a real solution may not exist.
4. Nonlinearity: The differential equation may be nonlinear, meaning that it involves higher order terms or products of the function and its derivatives. Nonlinear differential equations can be more difficult to solve and may not always have real solutions.
These are just a few possible reasons, and the specific reason for the non-existence of a real solution would depend on the specific form of the differential equation and its associated conditions.
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What is the reason behind the non-existence of any real function which satisfies the differential equation, (y’)2 + 1 = 0?a)Because for any real function, the left-hand side of the equation will be greater than, or equal to one and thus cannot be zerob)Because for any real function, the left-hand side of the equation will be less than, or equal to one and thus cannot be zeroc)Because for any real function, the left-hand side of the equation becomes zerod)Because for any real function, the left-hand side of the equation becomes infinityCorrect answer is option 'A'. Can you explain this answer?
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What is the reason behind the non-existence of any real function which satisfies the differential equation, (y’)2 + 1 = 0?a)Because for any real function, the left-hand side of the equation will be greater than, or equal to one and thus cannot be zerob)Because for any real function, the left-hand side of the equation will be less than, or equal to one and thus cannot be zeroc)Because for any real function, the left-hand side of the equation becomes zerod)Because for any real function, the left-hand side of the equation becomes infinityCorrect answer is option 'A'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about What is the reason behind the non-existence of any real function which satisfies the differential equation, (y’)2 + 1 = 0?a)Because for any real function, the left-hand side of the equation will be greater than, or equal to one and thus cannot be zerob)Because for any real function, the left-hand side of the equation will be less than, or equal to one and thus cannot be zeroc)Because for any real function, the left-hand side of the equation becomes zerod)Because for any real function, the left-hand side of the equation becomes infinityCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for What is the reason behind the non-existence of any real function which satisfies the differential equation, (y’)2 + 1 = 0?a)Because for any real function, the left-hand side of the equation will be greater than, or equal to one and thus cannot be zerob)Because for any real function, the left-hand side of the equation will be less than, or equal to one and thus cannot be zeroc)Because for any real function, the left-hand side of the equation becomes zerod)Because for any real function, the left-hand side of the equation becomes infinityCorrect answer is option 'A'. Can you explain this answer?.
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