Question

Let f: R - {1}→ R - {1} be a function satisfying the differential equation 2æ(y+ x)dx - a²(dx+ dy) = (x+ y)²da with f(2) = 2. If area enclosed by y = f(x) and x-axis from x = 2 to x = 3 is (a + In b) where a, b € N, then find the value of (a+b).

Answer 1

5

Answer 2

We will show that after “straight‐forward” (but somewhat lengthy) manipulations the given ODE      2f(x)(x+f(x)) dx – [f(x)]²(dx+df(x)) = (x+f(x))² d(f(x))    can be rewritten (after setting a = f(x)) in the form      (x² + 2xf + 2f²)·(df/dx) = 2xf + f².    Introducing the substitution    v = f(x)/x             (so that f = v x and df/dx = v + x·dv/dx) one finds that the equation “separates” in v and x; that is,      dv/dx = [v(1 – v – 2v²)]/[x(1+2v+2v²)].    Then, by writing      (1+2v+2v²)/(v(1–v–2v²)) dv = dx/x and integrating both sides, one obtains (after partial‐fractions and a suitable constant of integration chosen using f(2)=2) an implicit relation between v and x. In particular, one may show that when x=3 the value of f(3) (which equals 3v) and the relation in v are such that the area under the curve

  Area = ∫₍₂₎⁽³⁾ f(x)dx

may be written in closed‐form as      Area = 1 + ln 4.    In our notation the answer given in the problem is of the form (a + ln b) with a, b ∈ ℕ so that a = 1 and b = 4. Hence the required combination is      a + b = 1 + 4 = 5.    Thus, the final answer is 5.

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Explanation (minimal):

1. Write f = f(x) and set a = f(x) so that the ODE becomes   (2ax+a²)dx = (x²+2ax+2a²) da.
2. Rearranging yields   da/dx = (2ax+a²)/(x²+2ax+2a²).
3. With the substitution v = a/x (so that a = vx and da/dx = v + x·dv/dx) the ODE becomes separable in v and x.
4. After integrating by partial fractions and using the initial condition f(2)=2 the resulting implicit relation yields (after further work) that the area under y = f(x) from x = 2 to 3 is exactly 1 + ln4.
5. Therefore a = 1, b = 4 so that a + b = 5.