Question

Let the solution curve y = y(x) of the differential equation (4 + x²) dy – 2x (x² + 3y + 4) dx = 0 pass through the origin. Then y (2) is equal to ______.

Options

• 11

• 12

• 13

• 14

Answer 1

Let the solution curve y = y(x) of the differential equation (4 + x²) dy – 2x (x² + 3y + 4) dx = 0 pass through the origin. Then y (2) is equal to 12.

Explanation:

Given differential equation is

(4 + x²) dy – 2x (x² + 3y + 4) dx = 0

`\implies (x^2 + 4)(dy)/(dx)` = 2x³ + 6xy + 8x

`(x^2 + 4)(dy)/(dx) - 6xy` = 2x³ + 8x

`(dy)/(dx) - (6x)/(x^2 + 4)y = (2x^3 + 8x)/(x^2 + y)`

It looks like 

`(dy)/(dx) + py` = Q

where

p = `(-6x)/(x^2 + 4)`; Q = `(2x^3 + 8x)/(x^2 + 4)`

I.F. = `e^(-int (6x)/(x^2 + 4)dx) = e^(-3log_e(x^2 + 4)`

= `e^(log_e(x^2 + 4)^-3) = 1/(x^2 + 4)^3`

∴ `y. 1/(x^2 + 4)^3 = int  (2x^3 + 8x)/((x^2 + 4)^3(x^2 + 4))dx`

`y/(x^2 + 4)^3 = int (2x(x^2 + 4))/((x^2 + 4)^3(x^2 + 4))dx`

x² + 4 = t

2xdx = dt

`y/(x^2 + 4)^3 = int(dy)/t^3; y/(x^2 + 4)^3 = (-1)/(2(x^2 + 4)^2) + C`

Passes through origin (0, 0)

0 = `(-1)/(2 xx 16) + C; y/(x^2 + 4)^3 = (-1)/(2(x^2 + 4)^2) + 1/32`

y = `(-(x^2 + 4))/2 + (x^2 + 4)^3/32`

y(2) = `-8/2 + (8 xx 8 xx 8)/32`

= 16 – 4

= 12