Solve the given initial value problem correctly

Question

y(x)=

Answer 1

This is homogenous equation, substitution y = vx; y' = v + xv';

y' = y/x - 3x²/y²; v + xv' = v - 3/v²; xdv/dx = - 3/v²; v²dv = - 3dx/x; ∫v²dv = - 3∫dx/x; v³ = - 9lnx + C;

y³/x³ = - 9lnx + C; 8/1 = - 9ln1 + C; C = 8; y³ = - 9x³lnx + 8x³

Answer 2

ANOTHER APPROACH ; THE BERNOULLI

y² y' - (1/x) y³ = -3 x²

Let w = y³ then d w/ d x = 3 y² d y/ d x ⇒ y² d y/ d x = (1/3) d w/ d x Therefore

(1/3) d w/ d x - (1/x) w = -3 x²

d w/ d x - (3/x) w = -9 x² An integrating factor μ(x) = e^(-∫3/dx) = x^-3

x^-3 d w/ d x - (3/x^-2) w = -9 /x

d( x^-3 w ) = -9dx /x

∫d( x^-3 w ) = -9dx /x

x^-3 w = - 9lnx +ξ , ξ any real

y³ = -9 x³ ln x + ξ x³ and since y ( 1) =2 then ξ=8 and finally

y³ = -9 x³ ln x + 8 x³

ANOTHER APPROACH : IN SEARCH OF AN INTEGRATING FACTOR

The given differential equation can be written in the form.

( y³ - 3x³ ) d x - x y² d y = 0

Let M (x, y ) = y³ - 3x³ and N (x, y ) = - x y² Then

[ ∂M/ ∂y - ∂N / ∂x ] / N = - 4/ x The an integrating factor μ ( x ) = e^( - ∫4/x d x ) = 1 / x⁴

Multiplying the equation by 1 / x⁴ the equation becomes exact.

( y³/ x⁴ - 3/ x ) d x - y² / x³ d y = 0 ( E q 1.)

Then we conclude that Eq 1. is the total differential of Φ (x, y ) = c , where c constant.

Therefore ∂Φ/ ∂x = y³/ x⁴ - 3/ x ⇒ Φ (x, y ) = ∫ ( y³/ x⁴ - 3/ x ) d x + f (y) ⇒ Φ (x, y ) = - y³/ 3x³ - 3ln x + f (y)

Also since E q 1 is an exact one ∂Φ/ ∂y = [∂/∂y]{ - y³/ 3x³ - 3ln x + f (y) } = - y² / x³ + f ' (y)

Which is f ' (y) = 0 ⇒ f (y) = constant. y³ = -9 x³ ln x + ξ x³ and since y ( 1) =2 then ξ=8 and finally

Hence - y³/ 3x³ - 3ln x = ζ ⇒ y³ = -9 x³ ln x + 3ζx³ ⇒ y³ = -9 x³ ln x + ξ x³ and since y ( 1) =2 then ξ=8

and finally y³ = -9 x³ ln x + 8 x³

THIRD APPROACH : BY INSPECTION

The differential equation can be written in the form

y³ d x - x y² d y = 3x³ d x

y² [ x d y - y d x] = - 3x³ d x

3 (y²/ x² ){ [ x d y - y d x ] / x² } = -( 9/ x )d x

3 (y²/ x² )d ( y /x ) = -( 9/ x )d x

y³ / x³ = - 9 ln x + ξ

Then see solution one.