Question

In the solution to problem 1H-3 b), I don't understand why it is -x^2 instead of +x^2. Typo?

Solution to problem 1h-3 c): WolframAlpha disagrees with the solution: http://www.wolframalpha.com/input/?i=solve%28y^2-e^x*y-e^x%3D0%2Cy%29. The e^x term is missing from the OCW solution.

Answer 1

yes, for 1H-3 (b), the exponent should be +x^2. In their solution, they got off to a bad start by inserting -x^2 at the very beginning.

for 1H-3 c) they have a typo
they get to the quadratic
\[ y^2 - y \exp(x) -\exp(x) = 0 \]
and we use the quadratic formula
\[ y = \frac{-b \pm \sqrt{b^2 -4ac}}{2a} \]
to get
\[ y = \frac{\exp(x) \pm \sqrt{\exp(2x) + 4 \exp(x)}}{2} \]
because we have ln(y) in the original equation, and log's don't operate on negative values (assuming we are working with real numbers), we select only the positive root, namely
\[ y = \frac{\exp(x) +\sqrt{\exp(2x) + 4 \exp(x)}}{2} \]

it may not be obvious that the second root is negative... but as a test case notice
\[ \exp(0) - \sqrt{ \exp(0)+4 \exp(0)} = 1-\sqrt{5} \]
and that is negative, showing we must exclude this root.