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Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = ex sin x.
- a)4y − 1 = ex ( sin x − cos 2x)
- b)3y − 1 = ex ( sin x − cos2x)
- c)2y − 1 = ex ( sin x − cos x)
- d)2y + 1 = ex ( sin2 x − cos x)
Correct answer is option 'C'. Can you explain this answer?
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Find the equation of a curve passing through the point (0, 0) and whos...
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Find the equation of a curve passing through the point (0, 0) and whos...
Understanding the Differential Equation
The given differential equation is:
- y' = e^x * sin(x)
This indicates that we need to find a function y(x) whose derivative equals e^x * sin(x).
Integrating the Function
To find y, we integrate the right side with respect to x:
- y = ∫ e^x * sin(x) dx
To solve this integral, we can use integration by parts or the tabular method. Let's denote:
- u = sin(x) → du = cos(x) dx
- dv = e^x dx → v = e^x
Using integration by parts:
- ∫ e^x * sin(x) dx = e^x * sin(x) - ∫ e^x * cos(x) dx
Next, we apply integration by parts again on the remaining integral:
- Let:
- u = cos(x) → du = -sin(x) dx
- dv = e^x dx → v = e^x
This gives us:
- ∫ e^x * cos(x) dx = e^x * cos(x) + ∫ e^x * sin(x) dx
This leads to:
- ∫ e^x * sin(x) dx = e^x * sin(x) - (e^x * cos(x) + ∫ e^x * sin(x) dx)
This results in:
- 2 ∫ e^x * sin(x) dx = e^x * sin(x) - e^x * cos(x)
Solving for the integral yields:
- ∫ e^x * sin(x) dx = (1/2) * e^x * (sin(x) - cos(x))
Thus, we find:
- y = (1/2) * e^x * (sin(x) - cos(x)) + C
Applying the Initial Condition
Given that the curve passes through (0, 0):
- 0 = (1/2) * e^0 * (sin(0) - cos(0)) + C
- 0 = (1/2) * (0 - 1) + C
- C = 1/2
So the equation becomes:
- y = (1/2) * e^x * (sin(x) - cos(x)) + (1/2)
To eliminate the fraction, we multiply everything by 2:
- 2y = e^x * (sin(x) - cos(x)) + 1
- Rearranging gives us: 2y - 1 = e^x * (sin(x) - cos(x))
This matches option 'C':
- 2y - 1 = e^x * (sin(x) - cos(x))
Conclusion
Thus, the correct answer is indeed option 'C'.
The given differential equation is:
- y' = e^x * sin(x)
This indicates that we need to find a function y(x) whose derivative equals e^x * sin(x).
Integrating the Function
To find y, we integrate the right side with respect to x:
- y = ∫ e^x * sin(x) dx
To solve this integral, we can use integration by parts or the tabular method. Let's denote:
- u = sin(x) → du = cos(x) dx
- dv = e^x dx → v = e^x
Using integration by parts:
- ∫ e^x * sin(x) dx = e^x * sin(x) - ∫ e^x * cos(x) dx
Next, we apply integration by parts again on the remaining integral:
- Let:
- u = cos(x) → du = -sin(x) dx
- dv = e^x dx → v = e^x
This gives us:
- ∫ e^x * cos(x) dx = e^x * cos(x) + ∫ e^x * sin(x) dx
This leads to:
- ∫ e^x * sin(x) dx = e^x * sin(x) - (e^x * cos(x) + ∫ e^x * sin(x) dx)
This results in:
- 2 ∫ e^x * sin(x) dx = e^x * sin(x) - e^x * cos(x)
Solving for the integral yields:
- ∫ e^x * sin(x) dx = (1/2) * e^x * (sin(x) - cos(x))
Thus, we find:
- y = (1/2) * e^x * (sin(x) - cos(x)) + C
Applying the Initial Condition
Given that the curve passes through (0, 0):
- 0 = (1/2) * e^0 * (sin(0) - cos(0)) + C
- 0 = (1/2) * (0 - 1) + C
- C = 1/2
So the equation becomes:
- y = (1/2) * e^x * (sin(x) - cos(x)) + (1/2)
To eliminate the fraction, we multiply everything by 2:
- 2y = e^x * (sin(x) - cos(x)) + 1
- Rearranging gives us: 2y - 1 = e^x * (sin(x) - cos(x))
This matches option 'C':
- 2y - 1 = e^x * (sin(x) - cos(x))
Conclusion
Thus, the correct answer is indeed option 'C'.
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Question Description
Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = ex sin x.a)4y−1=ex(sinx−cos2x)b)3y−1=ex(sinx−cos2x)c)2y−1=ex(sinx− cosx)d)2y+1=ex(sin2x−cosx)Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = ex sin x.a)4y−1=ex(sinx−cos2x)b)3y−1=ex(sinx−cos2x)c)2y−1=ex(sinx− cosx)d)2y+1=ex(sin2x−cosx)Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = ex sin x.a)4y−1=ex(sinx−cos2x)b)3y−1=ex(sinx−cos2x)c)2y−1=ex(sinx− cosx)d)2y+1=ex(sin2x−cosx)Correct answer is option 'C'. Can you explain this answer?.
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