Science:Math Exam Resources/Courses/MATH103/April 2012/Question 02 (f)

MATH103 April 2012

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Question 02 (f)
Short answer question. Simplify your answer to the best possible. Show in detail how you arrive at your answer (work will be considered for this problem). Show that $y(x)={\frac {\ln(x)+2}{x}}$ is a solution to the differential equation $\displaystyle x^{2}y'+xy=1.$

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
You do not need to solve the differential equation to answer this question. You are just asked to show that the given function is a solution.

Hint 2
Try finding $y^{\prime }$ using the quotient rule and plugging $y^{\prime }$ and $y$ into the expression $x^{2}y^{\prime }+xy$. If you can show that this expression is equal to $1$, you will have shown that $y$ is a solution to the differential equation.

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. In order to see if $y(x)$ satisfies the differential equation, we need to find $y'$, substitute $y$ and $y'$ into the left side of the differential equation, and see if the resulting expression is equal to 1. Thus, first, we calculate $y'$. $y'(x)={\frac {x\left({\frac {1}{x}}+0\right)-(\ln(x)+2)}{x^{2}}}={\frac {-1-\ln(x)}{x^{2}}}$ Then we plug $y$ and $y'$ into the left hand side of the differential equation. ${\begin{aligned}x^{2}y'+xy&=x^{2}\left({\frac {-1-\ln(x)}{x^{2}}}\right)+x\left({\frac {\ln(x)+2}{x}}\right)\\&=-1-\ln(x)+\ln(x)+2=1\checkmark \end{aligned}}$ Because this equals the right hand side of the original differential equation, the solution $y(x)$ is correct.

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