Khan Academy Differential Equations and Linear Algebra

Khan Academy is another free resource for students who want another point of view on the same topics covered in our textbook and by bob in class.

You have to do a free registration to "begin learning" at the website:  https://www.khanacademy.org/

You can also go to You Tube Khan Academy and look at their playlists:
https://www.youtube.com/user/khanacademy/playlists

You can look there in You Tube (its own search window) for Differential Equations or Linear Algebra, but you may be overwhelmed by the long playlists, or you can try Googling them directly. For example, using the You Tube search with "Khan Academy Alternate Coordinate Systems" leads to the video"Change of Basis" which is first in this category:
https://www.youtube.com/watch?v=uvgru6FcyxU

Once you are logged in to the KA website which is more structured and lists the consecutive topics under each main topic, you can choose from two topics related to this course:

• Differential Equations:
  First order DEs [stop at Euler's method, ignore Exact Equations and Homogeneous Equations]
  https://www.khanacademy.org/math/differential-equations/first-order-differential-equations
  Second Order DEs [forget Laplace Transforms]:
  https://www.khanacademy.org/math/differential-equations/second-order-differential-equations
  
• Linear Algebra:
  Vectors and Spaces
  Matrix Transformations
  Alternate Coordinate Systems (bases) [Skip down to Change of Basis, then Eigen Everything]
  https://www.khanacademy.org/math/linear-algebra
  For our purposes Matrix Transformations has too much that we do not need---a matrix transformation is simply the mapping of every vector in an R^n space to a new vector in that same space by multiplying by a square matrix.

Some students love this, but not everyone. It depends on your learning style. It is also important to INTERACT with a human being actively not just listen passively. Other students, university math center help, your professor bob are all avenues for you to explore.