First Midterm Practice Test

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This is a practice exam for the first midterm. Feel free to drop any comment or question below. I will try to answer here as many as possible, until the day before the exam. This is a good opportunity to compare notes and work with other students. Enjoy!

  1. Prove that the function y=Ce^x is a solution of the equation y'-y=0 and find the particular solution satisfying y(0)=-1.
  2. Find the general solution of the equation 3e^x \tan y + (2-e^x)\sec^2 y\, \frac{dy}{dx} =0.
  3. Find the particular solution of the equation (1+e^x)yy'=e^x that satisfies the initial condition y(0)=1.
  4. Find the equation of a curve that goes through the point (0, -2) and satisfies that the slope at any of its points is equal to three plus the y-coordinate at that point.
  5. Find the general solution of the equation xy' = \sqrt{x^2-y^2} + y.
  6. Find the general solution of the equation (x+y-2) + (x-y+4) y' =0.
  7. Find the general solution of the equation (x+y+1) + (2x+2y-1) y' = 0.
  8. Find the general solution of the equation y'+2xy=2xe^{-x^2}.
  9. Find the general solution of the equation xy'+y=y^2\ln x.
  10. Find the general solution of the second-order differential equation y''=0.
  11. Find the particular solution to the equation y''=2y^3 that satisfies the initial conditions y(0)=1 and y'(0)=1.
  12. And let’s finish with a nice punch-line: Find the general solution of the equation y''+(y')^2=2e^{-y}. If you are able to get this question in less than 30 minutes without the help of a computer, and explain to someone else step-by-step how to do it, I consider that you have mastered the material of the first midterm.
  1. Warren Hart
    January 29, 2012 at 9:36 pm
    Reply

    How would you solve #5?

    • Francisco Blanco-Silva
      January 29, 2012 at 9:42 pm
      Reply

      What did you try, and where did you get stuck? I can help you with hints from there.

      • Warren Hart
        January 29, 2012 at 9:55 pm
        Reply

        I am studying with my friend Cam, we think its a Bernoulli’s equation but we don’t really know how to solve it from there, we have tried to separate it but it doesn’t work, as well as we have tried the homogeneous route but that didn’t work either, our main problem is just solving a Bernoulli’s equation and because that is where we start we are pretty much screwed…

        • Francisco Blanco-Silva
          January 29, 2012 at 9:58 pm
          Reply

          It is homogeneous. Write the equation in the form y'=f(x,y) and do proper algebraic simplifications to realize it.

  2. Warren Hart
    January 29, 2012 at 10:09 pm
    Reply

    Ok I think I see what you are talking about now…thankyou.

  3. Sara Davino
    January 30, 2012 at 10:05 pm
    Reply

    Could you post what the final answers are so that I can check to see if my work is right? Thanks!

    • Francisco Blanco-Silva
      January 30, 2012 at 11:55 pm
      Reply

      No, sorry. I will definitely drop hints in the right direction, but no solutions.

  4. Samruddhi Somanii
    January 31, 2012 at 11:35 am
    Reply

    I have a question about 11. I substitute p=y', which led to a separable equation with the solution p^2=y^4+2c. Substituting back in y' gives (y')^2=y^4+2c. Thus, \frac{dy}{dx}=\sqrt{y^4+2c}. How would I go about solving this part? Thank you.

    • Francisco Blanco-Silva
      January 31, 2012 at 2:13 pm
      Reply

      It looks like an autonomous equation (separable without x). Before you integrate it, use the initial conditions, and see if you can get rid of the constant in your expression. Get back to me if you can’t figure it out.

      • Samruddhi Somani
        January 31, 2012 at 5:18 pm
        Reply

        I’m still not quite sure what to do in order to solve for the constant.

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