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If y1 = sinx and y2 = sinx – cosx are linearly independent solutions of y” + y = 0. Then determine the constants c1 and c2 so that the solution sin x + 3cosx = c1y1 + c2y2.
- a)c1 = 3, c2 = 4
- b)c1 = 4, c2 = 3
- c)c1 = –3, c2 = 4
- d)c1 = 4, c2 = –3
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
If y1 = sinx and y2 = sinx – cosx are linearly independent solut...
Given that the solutions of the differential equation y'' + y = 0 are y1 = sinx and y2 = sinx cosx, we need to determine the constants c1 and c2 such that the solution sinx 3cosx can be written as c1y1 + c2y2.
To find c1 and c2, we can use the fact that the Wronskian of two linearly independent solutions of a second-order linear homogeneous differential equation is a constant.
The Wronskian of y1 and y2 is given by:
W(y1, y2) = y1y2' - y1'y2
Differentiating y1 = sinx with respect to x, we get:
y1' = cosx
Differentiating y2 = sinx cosx with respect to x, we get:
y2' = cos^2(x) - sin^2(x)
Substituting these values into the Wronskian formula, we have:
W(y1, y2) = sinx(cos^2(x) - sin^2(x)) - cosx(sinx cosx)'
= sinx(cos^2(x) - sin^2(x)) - cosx(cos^2(x) - sin^2(x) - sinx cosx)
= sinx(cos^2(x) - sin^2(x)) - cosx(cos^2(x) - sin^2(x)) - cos^2(x)sinx + sin^3(x)
= 0
Since the Wronskian is zero, this implies that y1 and y2 are linearly dependent.
Now, let's express sinx 3cosx as c1y1 + c2y2 and solve for c1 and c2.
sinx 3cosx = c1y1 + c2y2
sinx 3cosx = c1sinx + c2sinx cosx
sinx 3cosx = (c1 + c2)sinx + c2sinx cosx
Comparing the coefficients of sinx and sinx cosx on both sides, we get:
c1 + c2 = 1 (coefficient of sinx)
c2 = 3 (coefficient of sinx cosx)
Therefore, the values of c1 and c2 are c1 = 1 - c2 = 1 - 3 = -2 and c2 = 3.
Hence, the correct answer is option D: c1 = 4, c2 = 3.
To find c1 and c2, we can use the fact that the Wronskian of two linearly independent solutions of a second-order linear homogeneous differential equation is a constant.
The Wronskian of y1 and y2 is given by:
W(y1, y2) = y1y2' - y1'y2
Differentiating y1 = sinx with respect to x, we get:
y1' = cosx
Differentiating y2 = sinx cosx with respect to x, we get:
y2' = cos^2(x) - sin^2(x)
Substituting these values into the Wronskian formula, we have:
W(y1, y2) = sinx(cos^2(x) - sin^2(x)) - cosx(sinx cosx)'
= sinx(cos^2(x) - sin^2(x)) - cosx(cos^2(x) - sin^2(x) - sinx cosx)
= sinx(cos^2(x) - sin^2(x)) - cosx(cos^2(x) - sin^2(x)) - cos^2(x)sinx + sin^3(x)
= 0
Since the Wronskian is zero, this implies that y1 and y2 are linearly dependent.
Now, let's express sinx 3cosx as c1y1 + c2y2 and solve for c1 and c2.
sinx 3cosx = c1y1 + c2y2
sinx 3cosx = c1sinx + c2sinx cosx
sinx 3cosx = (c1 + c2)sinx + c2sinx cosx
Comparing the coefficients of sinx and sinx cosx on both sides, we get:
c1 + c2 = 1 (coefficient of sinx)
c2 = 3 (coefficient of sinx cosx)
Therefore, the values of c1 and c2 are c1 = 1 - c2 = 1 - 3 = -2 and c2 = 3.
Hence, the correct answer is option D: c1 = 4, c2 = 3.
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Community Answer
If y1 = sinx and y2 = sinx – cosx are linearly independent solut...
As it's given y1=sinx and y2=sinx-cosx,
also given that sinx+3cosx= c1y1+c2y2
so sinx +3cosx = c1sinx+C2(sinx-cosx)
so sinx+3cosx=(C1+C2)sinx-c2cosx
so c2=-3 and c1+c2=1
soc1=4
also given that sinx+3cosx= c1y1+c2y2
so sinx +3cosx = c1sinx+C2(sinx-cosx)
so sinx+3cosx=(C1+C2)sinx-c2cosx
so c2=-3 and c1+c2=1
soc1=4
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If y1 = sinx and y2 = sinx – cosx are linearly independent solutions of y” + y = 0. Then determine the constants c1 and c2 so that the solution sin x + 3cosx = c1y1 + c2y2.a)c1 = 3, c2 = 4b)c1 = 4, c2 = 3c)c1 = –3, c2 = 4d)c1 = 4, c2 = –3Correct answer is option 'D'. Can you explain this answer? for Physics 2025 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about If y1 = sinx and y2 = sinx – cosx are linearly independent solutions of y” + y = 0. Then determine the constants c1 and c2 so that the solution sin x + 3cosx = c1y1 + c2y2.a)c1 = 3, c2 = 4b)c1 = 4, c2 = 3c)c1 = –3, c2 = 4d)c1 = 4, c2 = –3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Physics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If y1 = sinx and y2 = sinx – cosx are linearly independent solutions of y” + y = 0. Then determine the constants c1 and c2 so that the solution sin x + 3cosx = c1y1 + c2y2.a)c1 = 3, c2 = 4b)c1 = 4, c2 = 3c)c1 = –3, c2 = 4d)c1 = 4, c2 = –3Correct answer is option 'D'. Can you explain this answer?.
If y1 = sinx and y2 = sinx – cosx are linearly independent solutions of y” + y = 0. Then determine the constants c1 and c2 so that the solution sin x + 3cosx = c1y1 + c2y2.a)c1 = 3, c2 = 4b)c1 = 4, c2 = 3c)c1 = –3, c2 = 4d)c1 = 4, c2 = –3Correct answer is option 'D'. Can you explain this answer? for Physics 2025 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about If y1 = sinx and y2 = sinx – cosx are linearly independent solutions of y” + y = 0. Then determine the constants c1 and c2 so that the solution sin x + 3cosx = c1y1 + c2y2.a)c1 = 3, c2 = 4b)c1 = 4, c2 = 3c)c1 = –3, c2 = 4d)c1 = 4, c2 = –3Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for Physics 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If y1 = sinx and y2 = sinx – cosx are linearly independent solutions of y” + y = 0. Then determine the constants c1 and c2 so that the solution sin x + 3cosx = c1y1 + c2y2.a)c1 = 3, c2 = 4b)c1 = 4, c2 = 3c)c1 = –3, c2 = 4d)c1 = 4, c2 = –3Correct answer is option 'D'. Can you explain this answer?.
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