Science:Math Exam Resources/Courses/MATH105/April 2013/Question 03 (a)

MATH105 April 2013

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Question 03 (a)
Solve the following initial value problem: ${\frac {dy}{dx}}={\frac {1}{(x^{2}+x)y}}\quad y(1)=2.$

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 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
What kind of differential equation is given in this initial value problem? Which method should you use to solve it?

Hint 2
This is a separable differential equation. Isolate the terms with y and with x on separate sides. Then integrate.

Hint 3
For the more complicated integral, use partial fractions.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

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. First, we recognize that the differential equation given in this initial value problem is a separable differential equation. Hence, we move all terms involving $x,y$ to opposite sides of the equal sign and take the indefinite integral: $\int y\,dy=\int {\frac {dx}{x^{2}+x}}.$ The integral on the left hand side is easy, but some work is required to solve the integral on the right hand side. First, we rewrite it as ${\begin{aligned}\int y\,dy&=\int {\frac {1}{x(x+1)}}\,dx\\\end{aligned}}$ We proceed by using partial fractions to split up the integral. Notice that the fraction can be split up as ${\begin{aligned}{\frac {1}{x(x+1)}}=\left({\frac {A}{x}}+{\frac {B}{x+1}}\right)={\frac {A(x+1)+B(x)}{x(x+1)}}.\end{aligned}}$ This gives ${\begin{aligned}1=A(x+1)+B(x)\end{aligned}}$ Setting x to 0 gives $A=1$ and setting x to -1 gives $B=-1$. Thus, we can write out integral as ${\begin{aligned}\int y\,dy&=\int {\frac {1}{x(x+1)}}\,dx\\\int y\,dy&=\int \left({\frac {1}{x}}-{\frac {1}{x+1}}\right)\,dx.\end{aligned}}$ Now we integrate both sides of the equation and add the arbitary constant, $C$, to the right hand side: ${\frac {1}{2}}y^{2}=\ln |x|-\ln |x+1|+C.$ Isolating for $y$ we get: $y(x)=\pm {\sqrt {2\ln |x|-2\ln |x+1|+K}}.$ where $K=2C$. Since the initial condition, $y(1)=2>0$, we take the positive solution for $y$: $y(x)={\sqrt {2\ln |x|-2\ln |x+1|+K}}.$ To solve for $K$, we apply the initial condition to the solution: ${\begin{aligned}y(1)={\sqrt {2\ln |1|-2\ln |2|+K}}&=2\\{\sqrt {K-2\ln(2)}}&=2\\K-2\ln(2)&=4\\K&=2\ln(2)+4.\end{aligned}}$ Therefore, the solution to the initial value problem is: ${\color {blue}y(x)={\sqrt {2\ln |x|-2\ln |x+1|+2\ln(2)+4}}.}$

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