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The value of y at x = 0.1 to five places of decimals, by Taylor's series method, given that dy/dx = x2y−1, y(0) = 1, is
- a)0.68281
- b)0.81122
- c)0.90033
- d)0.70127
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The value of y at x = 0.1 to five places of decimals, by Taylors serie...
To find the value of y at x = 0.1 using Taylor series method, we need to find the coefficients of the Taylor series expansion of y.
The given differential equation is dy/dx = x^2y.
Let's differentiate both sides of the equation with respect to x:
d^2y/dx^2 = 2xy + x^2(dy/dx).
Now, let's substitute x = 0 into the differential equation:
d^2y/dx^2 = 0^2 * y = 0.
This means that the second derivative of y with respect to x is zero at x = 0.
We can use this information to write the Taylor series expansion of y as follows:
y = y(0) + x(dy/dx)(0) + (x^2/2!)(d^2y/dx^2)(0) + ...
Since the second derivative is zero at x = 0, the Taylor series expansion simplifies to:
y = y(0) + x(dy/dx)(0) + ...
Now, let's find the values of y(0) and (dy/dx)(0) using the given differential equation.
At x = 0, dy/dx = 0^2 * y = 0.
So, (dy/dx)(0) = 0.
Now, let's differentiate the differential equation with respect to x again:
d^2y/dx^2 = 2xy + x^2(dy/dx).
Substituting x = 0, we get:
d^2y/dx^2 = 0.
Therefore, the second derivative is also zero at x = 0, which means (d^2y/dx^2)(0) = 0.
Now, we can write the Taylor series expansion as:
y = y(0) + 0 + ...
Since we only need the value of y at x = 0.1 to five decimal places, we can stop here.
Therefore, the value of y at x = 0.1 is equal to y(0).
However, without specific initial conditions or more information, we cannot determine the exact value of y(0) or the value of y at x = 0.1 using the Taylor series method.
The given differential equation is dy/dx = x^2y.
Let's differentiate both sides of the equation with respect to x:
d^2y/dx^2 = 2xy + x^2(dy/dx).
Now, let's substitute x = 0 into the differential equation:
d^2y/dx^2 = 0^2 * y = 0.
This means that the second derivative of y with respect to x is zero at x = 0.
We can use this information to write the Taylor series expansion of y as follows:
y = y(0) + x(dy/dx)(0) + (x^2/2!)(d^2y/dx^2)(0) + ...
Since the second derivative is zero at x = 0, the Taylor series expansion simplifies to:
y = y(0) + x(dy/dx)(0) + ...
Now, let's find the values of y(0) and (dy/dx)(0) using the given differential equation.
At x = 0, dy/dx = 0^2 * y = 0.
So, (dy/dx)(0) = 0.
Now, let's differentiate the differential equation with respect to x again:
d^2y/dx^2 = 2xy + x^2(dy/dx).
Substituting x = 0, we get:
d^2y/dx^2 = 0.
Therefore, the second derivative is also zero at x = 0, which means (d^2y/dx^2)(0) = 0.
Now, we can write the Taylor series expansion as:
y = y(0) + 0 + ...
Since we only need the value of y at x = 0.1 to five decimal places, we can stop here.
Therefore, the value of y at x = 0.1 is equal to y(0).
However, without specific initial conditions or more information, we cannot determine the exact value of y(0) or the value of y at x = 0.1 using the Taylor series method.
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Community Answer
The value of y at x = 0.1 to five places of decimals, by Taylors serie...
Hence by Taylor series:
1 - 0.1 + 0 + 0.00033 + ......
= 0.90031 ≈ 0.90033
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The value of y at x = 0.1 to five places of decimals, by Taylors series method, given thatdy/dx = x2y−1, y(0) = 1, isa)0.68281b)0.81122c)0.90033d)0.70127Correct answer is option 'C'. Can you explain this answer?
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The value of y at x = 0.1 to five places of decimals, by Taylors series method, given thatdy/dx = x2y−1, y(0) = 1, isa)0.68281b)0.81122c)0.90033d)0.70127Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The value of y at x = 0.1 to five places of decimals, by Taylors series method, given thatdy/dx = x2y−1, y(0) = 1, isa)0.68281b)0.81122c)0.90033d)0.70127Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The value of y at x = 0.1 to five places of decimals, by Taylors series method, given thatdy/dx = x2y−1, y(0) = 1, isa)0.68281b)0.81122c)0.90033d)0.70127Correct answer is option 'C'. Can you explain this answer?.
The value of y at x = 0.1 to five places of decimals, by Taylors series method, given thatdy/dx = x2y−1, y(0) = 1, isa)0.68281b)0.81122c)0.90033d)0.70127Correct answer is option 'C'. Can you explain this answer? for Civil Engineering (CE) 2025 is part of Civil Engineering (CE) preparation. The Question and answers have been prepared according to the Civil Engineering (CE) exam syllabus. Information about The value of y at x = 0.1 to five places of decimals, by Taylors series method, given thatdy/dx = x2y−1, y(0) = 1, isa)0.68281b)0.81122c)0.90033d)0.70127Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The value of y at x = 0.1 to five places of decimals, by Taylors series method, given thatdy/dx = x2y−1, y(0) = 1, isa)0.68281b)0.81122c)0.90033d)0.70127Correct answer is option 'C'. Can you explain this answer?.
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