Science:Math Exam Resources/Courses/MATH215/December 2011/Question 02 (a)

MATH215 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q7 (d)  •  

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Question 02 (a)
Find all values of a and b for which the ODE below is an exact equation. (It is not necessary to solve the ODE.) Failed to parse (syntax error): {\displaystyle ax^3e^{x+y} + bx^4e^{x+y} + 2x + (x^4e^{x+y} + 2y) \frac{dy}{dx} = 0 }

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
An differential equation is exact if it can be derived by differentiating an implicit equation $M(x,y)=0$ with respect to x. After differentiating with respect to x, think about Clairaut's theorem...

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We begin with assuming there is a governing equation $M(x,y)=0$ and differentiating this with respect to x. Then, ${\frac {d}{dx}}M(x,y)=M_{x}(x,y)+M_{y}(x,y){\frac {dy}{dx}}=0$. Hence, $M_{x}=ax^{3}e^{x+y}+bx^{4}e^{x+y}+2x$ and $M_{y}=x^{4}e^{x+y}+2y$. By Clairaut's theorem, there must be equality of mixed partial derivatives, and $M_{xy}=M_{yx}$. In our problem, ${\begin{aligned}M_{xy}&=ax^{3}e^{x+y}+bx^{4}e^{x+y}\\M_{yx}&=4x^{3}e^{x+y}+x^{4}e^{x+y}\\&\implies a=4,b=1\end{aligned}}$ by comparing the coefficients of $x^{3}e^{x+y}$ and $x^{4}e^{x+y}$.

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