 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-2/14/b/i/solution

With

v = x + y

, the chain rule gives

\frac{d v}{d x} = 1 + \frac{d y}{d x} = \frac{2 ( v - 2 )}{3 v - 4} .

For

v \neq = 2

x = \int \frac{3 v - 4}{2 ( v - 2 )} d v = \frac{3}{2} v + lo g ∣ v - 2∣ + C .

Restoring

v

and absorbing constants, the nonconstant branches satisfy

x + 3 y + 2 lo g ∣ x + y - 2∣ = C .

The constant branch lost in division is

v = 2

, giving the additional solution

y = 2 - x

. All branches must avoid

v = 4/3

, where the original differential equation is undefined.

 New to topics? Read the docs here!