Science:Math Exam Resources/Courses/MATH101/April 2016/Question 04 (a)

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MATH101 April 2016

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Question 04 (a)
Find the function $y=y(x)$ that satisfies $y(1)=4$ and ${\frac {dy}{dx}}={\frac {15x^{2}+4x+3}{y}}.$

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
Try separation of variables method.

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Solution
We separate the variables (cross multiply to have $y$ at one side, $x$ at the other side): $ydy=(15x^{2}+4x+3)dx$ Now integrate: ${\begin{aligned}\int ydy&=\int (15x^{2}+4x+3)dx\\{\frac {y^{2}}{2}}&=5x^{3}+2x^{2}+3x+\color {red}{C}.\end{aligned}}$ Plugging in $x=1$ and $y=4$ gives: ${\frac {4^{2}}{2}}=5+2+3+C,$ and so $\color {red}{C=-2}$ . This implies ${\frac {y^{2}}{2}}=5x^{3}+2x^{2}+3x-2$ . or $y=$ ± ${\sqrt {10x^{3}+4x^{2}+6x-4}}.$ Since $y(1)=4$ is positive, so we should choose the positive square root $\color {blue}{y={\sqrt {10x^{3}+4x^{2}+6x-4}}}.$ .