Science:Math Exam Resources/Courses/MATH101/April 2018/Question 06 (i)

MATH101 April 2018

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Question 06 (i)
Suppose that ${\textstyle y(x)}$ satisfies the differential equation $${\frac {dy}{dx}}=-{\frac {2\sin(x)}{y^{3}}}\,,\qquad {\mbox{with}}\quad y(0)=-2\,.$$ For this solution, calculate ${\textstyle y({\pi /2})}$.

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
Notice that the given equation is a separable differential equation.

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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. Since the given equation is a separable differential equation, we first collect the function with ${\textstyle y}$ variable on the left side of the equation and the one with ${\textstyle x}$ variable on the right, then integrates both side: $$\int y^{3}dy=-\int 2\sin(x)dx$$ By the power rule, the integral on the left side can be evaluated as $$\int y^{3}dy={\frac {1}{4}}y^{4}+C_{1}.$$ On the other hand, the right side one is $$-\int 2\sin(x)dx=2\cos(x)+C_{2}.$$ Here, ${\textstyle C_{1}}$ and ${\textstyle C_{2}}$ are arbitrary constants. Plugging these into the equation, we have $${\frac {1}{4}}y^{4}=2\cos(x)+C,$$ where ${\textstyle C=C_{2}-C_{1}}$ is also an arbitrary constant. To find ${\textstyle C}$, we use ${\textstyle y(0)=-2}$; $$C={\frac {1}{4}}y(0)^{4}-2\cos(0)={\frac {1}{4}}(-2)^{4}-2\cdot 1=4-2=2.$$ Therefore, we get $$y^{4}=8\cos(x)+8\implies y=\pm (8\cos(x)+8)^{\frac {1}{4}}.$$ Considering that $y(0)=-2$, we take $y$ with negative sign as the solution: $$y=-(8\cos(x)+8)^{\frac {1}{4}}.$$ Finally, plugging ${\textstyle x=\pi /2}$ into the solution to get $$y\left({\frac {\pi }{2}}\right)=-(8\cos \left({\frac {\pi }{2}}\right)+8)^{\frac {1}{4}}=-(2^{3})^{\frac {1}{4}}=-2^{\frac {3}{4}}.$$