JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Differential Equations

Let

y = y (x)

be the solution of the differential equation

(x - x^{3}) d y = (y + y x^{2} - 3 x^{4}) d x, x > 2

. If

y (3) = 3

then

y (4)

is equal to

Answer: (A)

Solution

Expanding the equation,

x d y - x^{3} d y = y d x + y x^{2} d x - 3 x^{4} d x

, that is

(x d y - y d x) - x^{2} (x d y + y d x) + 3 x^{4} d x = 0

Dividing throughout by

x^{2}

\frac{x d y - y d x}{x ^{2}} - (x d y + y d x) + 3 x^{2} d x = 0

and each group is an exact differential:

d (\frac{y}{x}) - d (x y) + d (x^{3}) = 0

Integrating gives

\frac{y}{x} - x y + x^{3} = c

. Putting

x = 3, y = 3

gives

1 - 9 + 27 = c

, so

c = 19

. At

x = 4

\frac{y}{4} - 4 y + 64 = 19 ⟹ - \frac{15 y}{4} = - 45 ⟹ y = 12

y (4) = 12

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