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Mid2Solution

Mid2Solution - Numerical Analysis in Engineering ME 140A...

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Numerical Analysis in Engineering ME 140A, Fall 2008 Midterm #2 Solution By: Mohamad M. Nasr-Azadani December 3, 2008 Problem 1 Using separation of variables, the given ODE can be re-arranged as y 2 dy = 2 dx x . (1) Integrating both sides of equation (1) will give us Z y 2 dy = Z 2 dx x y 3 3 = 2 ln | x | + C (2) Here, C is a constant which can be found using the initial condition. Equation (2) gives y as a function of x implicitly. Finally, y is found as y = (6 ln | x | + 3 C ) 1 3 . (3) 1

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For the ﬁrst order ODE with the form given by y 0 + p ( x ) = g ( x ) (4) the integrating factor μ ( x ) is found by μ ( x ) = e R p ( x ) dx . (5) Hence, we have μ ( x ) = e R 2 xdx = e x 2 (6) Multiplying μ ( x ) to the both sides of equation (4), we have d dx [ μy ] = g ( x ) μ ( x ) (7) Integrating both sides of equation (7) results in μ ( x ) y = Z g ( x ) μ ( x ) dx (8) Using equation (6), we have e x 2 y = Z xe x 2 dx = Z 1 2 (2 x ) e x 2 dx = 1 2 e x 2 + C
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Mid2Solution - Numerical Analysis in Engineering ME 140A...

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The text passage is a solution to a midterm exam problem in Numerical Analysis in Engineering, specifically dealing with solving ordinary differential equations (ODEs) using separation of variables and integrating factors.
no_question

the passage does not contain a mathematical question but rather provides a solution to a problem. it explains the steps to solve a given ode using separation of variables and integrating factors, and it does not provide an explicit question or a separate answer section.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

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TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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Mid2Solution

Mid2Solution - Numerical Analysis in Engineering ME 140A...

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Numerical Analysis in Engineering ME 140A, Fall 2008 Midterm #2 Solution By: Mohamad M. Nasr-Azadani December 3, 2008 Problem 1 Using separation of variables, the given ODE can be re-arranged as y 2 dy = 2 dx x . (1) Integrating both sides of equation (1) will give us Z y 2 dy = Z 2 dx x y 3 3 = 2 ln | x | + C (2) Here, C is a constant which can be found using the initial condition. Equation (2) gives y as a function of x implicitly. Finally, y is found as y = (6 ln | x | + 3 C ) 1 3 . (3) 1

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For the ﬁrst order ODE with the form given by y 0 + p ( x ) = g ( x ) (4) the integrating factor μ ( x ) is found by μ ( x ) = e R p ( x ) dx . (5) Hence, we have μ ( x ) = e R 2 xdx = e x 2 (6) Multiplying μ ( x ) to the both sides of equation (4), we have d dx [ μy ] = g ( x ) μ ( x ) (7) Integrating both sides of equation (7) results in μ ( x ) y = Z g ( x ) μ ( x ) dx (8) Using equation (6), we have e x 2 y = Z xe x 2 dx = Z 1 2 (2 x ) e x 2 dx = 1 2 e x 2 + C
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The text passage is a solution to a midterm exam problem in Numerical Analysis in Engineering, specifically dealing with solving ordinary differential equations (ODEs) using separation of variables and integrating factors.

TYPE: NO_QUESTION

The passage does not contain a mathematical question but rather provides a solution to a problem. It explains the steps to solve a given ODE using separation of variables and integrating factors, and it does not provide an explicit question or a separate answer section.