Answers to Problems in the Notes of Geometric Applications
Problem 1.
- \( y^2 = Cx + D. \)
- \( x^2 + y^2 = C. \)
- \( y^2 = Ce^x. \)
- \( y = C - x. \)
- \( \ln \lvert x \rvert = C - \tfrac{1}{x}. \)
- \( x^2 - y^2 = C. \)
- Two possible solutions: \( y^2 \pm 2y^2 = C. \)
- I suspect there is a typo in the statement of this problem, because this one is pretty tough. I got it reduced to the equation \( (x-y’-y)^2 (y’)^2= y^2 \big( 1+ (y’)^2 \big) \). We’d have to find all solutions of this equation for \( y’ \), and then solve each of those differential equations independently. ANGTFT
Problem 2.
- \( y = 2x+C. \)
- \( \lvert y \rvert = C \sqrt[n]{\lvert x \rvert}. \)
- \( y^2 = \tfrac{2}{n}x^2 + C. \)
- \( \lvert y \rvert = Cx^2. \)
- We’ll do this one later on. Focus on the ones for which we have discussed the method \( F(x,y) = C. \)
- \( y^2 = x + C. \)
- Pretty tough. \( \big( \tfrac{y^2}{x^2}+1 \big) \big( \tfrac{y^2}{x^2}+2 \big)^{-1/2} = Ce^{-x}. \)
- \( \tfrac{1}{y-x} - \ln \big( 1-\tfrac{1}{y-x} \big) = C-x. \)
- One solution is \( y = 0. \) The other is \( \tfrac{2}{3}x^3 + xy^2 = C. \)