Question

Let y = y(x) be the solution of the differential equation
$x ( 1 - x² ) \frac{dy}{dx} + ( 3x²y - y - 4x³ ) = 0, x > 1$
with y(2) = –2. Then y(3) is equal to

• A. -18
• B. -12
• C. -6
• D. -3

Answer 1

Answer: (A)

-18

Answer 2

The correct answer is (A) : -18
$\frac{dy}{dx} + \frac{y(3x^2 - 1)}{x(1 - x^2)} = \frac{4x^3}{x(1 - x^2)}$
$\text{IF} = e^{\int \frac{3x^2 - 1}{x - x^3} \,dx} = e^{-\ln |x^3 - x|} = e^{-\ln(x^3 - x)}$
$= \frac{1}{x³ - x}$
Solution of D.E. can be given by
$y \cdot \frac{1}{x^3 - x} = \int \frac{4x^3}{x(1 - x^2)} \cdot \frac{1}{x(x^2 - 1)} \,dx$
$⇒$ $$$\frac{y}{x^3 - x} = \int \frac{-4x}{(x^2 - 1)^2} ,dx$$⇒$$\frac{y}{x^3 - x} = \frac{2}{x^2 - 1} + c$at x = 2, y = -2$\frac{-2}{6} = \frac{2}{3} + c ⇒ c = -1$at$x = 3 ⇒ \frac{y}{24} = \frac{2}{8} - 1 ⇒ y = -18$