Science:Math Exam Resources/Courses/MATH101/April 2015/Question 09 (a)

MATH101 April 2015

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Question 09 (a)
(a) Find the solution of the differential equation ${\frac {1+{\sqrt {y^{2}-4}}}{\tan x}}y'={\frac {\sec x}{y}}$ that satisfies $y(0)=2$. You don’t have to solve for $y$ in terms of $x$.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
It is a separable differential equation.

Hint 2
Recall that $\tan x={\frac {\sin x}{\cos x}},\quad \sec x={\frac {1}{\cos x}}$.

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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 first separate the variables to have ${\begin{aligned}(1+{\sqrt {y^{2}-4}})yy'=\sec x\tan x&\Rightarrow (1+{\sqrt {y^{2}-4}})y\,dy=\sec x\tan x\,dx\\&\Rightarrow \int (1+{\sqrt {y^{2}-4}})y\,dy=\int \sec x\tan x\,dx+C\\\end{aligned}}$ Now we compute the integrals on each side: For the left-hand- side integral, we do the substitution: ${\sqrt {y^{2}-4}}=u\Rightarrow {\frac {y\,dy}{\sqrt {y^{2}-4}}}=du\Rightarrow y\,dy={\sqrt {y^{2}-4}}\,du=u\,du$ so: $\int (1+{\sqrt {y^{2}-4}})y\,dy=\int (1+u)u\,du=\int u+u^{2}\,du={\frac {1}{2}}u^{2}+{\frac {1}{3}}u^{3}={\frac {1}{2}}(y^{2}-4)+{\frac {1}{3}}({\sqrt {y^{2}-4}})^{3}$ For the right-hand side we have $\int \sec x\tan x\,dx=\int {\frac {\sin x}{\cos ^{2}x}}dx$ Now apply the following substitution $\cos x=u,-\sin x\,dx=du$ $\int {\frac {\sin x}{\cos ^{2}x}}dx=\int {\frac {-du}{u^{2}}}=\int -u^{-2}du=u^{-1}={\frac {1}{\cos x}}$ We computed the integrals on both sides, thus ${\frac {1}{2}}(y^{2}-4)+{\frac {1}{3}}({\sqrt {y^{2}-4}})^{3}={\frac {1}{\cos x}}+C$ The last step is to use the given initial condition to find the constant $C$: $y(0)=2$ gives: ${\frac {1}{2}}(2^{2}-4)+{\frac {1}{3}}({\sqrt {2^{2}-4}})^{3}={\frac {1}{\cos 0}}+C\Rightarrow 0=1+C\Rightarrow C=-1$ Therefore, ${\color {blue}{\frac {1}{2}}(y^{2}-4)+{\frac {1}{3}}({\sqrt {y^{2}-4}})^{3}={\frac {1}{\cos x}}-1}$