# classify the equation: dy/dx = x / (x^2 y + y^3) linear, nonlinear, separable,exact, homogeneous, or one that requires an integration factor?

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You notice that this is an exact differential equation:`dy/dx = x / (x^2 y + y^3)`

You need to multiply by dx both sides such that:

`dy = (xdx)/(x^2y + y^3)`

You need to multiply by `(x^2y + y^3)` such that:

`(x^2y + y^3)dy = xdx =gt (x^2y + y^3)dy- xdx = 0`

You should come up with the notation: the differential dy is multiplied by a function P(x,y).

You need to integrate the function P(x,y) with respect to y such that:`int (x^2y + y^3) dy= x^2y^2/2 + y^4/4` .

You should come up with the notation: the differential dx is multiplied by a function Q(x,y).

You need to integrate the function Q(x,y) with respect to x such that:`int x dx=x^2/2` .

Considering once each term from both integrals yields:

`f(x,y) = x^2y^2/2 + y^4/4 + x^2/2`

**Hence, the general solution to the exact differential equation is `x^2y^2/2 + y^4/4 + x^2/2 = c` .**

( dy／dx ) = x／( x2y + y3 ) → ( dx／dy ) = xy + ( y3／x )

→ ( dx／dy ) - yx = y3x - 1 ～ **Bernoulli's o.d.e.**

→ x ( dx／dy ) - yx2 = y3

Let u = x2 → ( du／dy ) = 2x ( dx／dy )

→ x ( dx／dy ) = ( 1／2 )( du／dy )

→ ( 1／2 )( du／dy ) - yu = y3

→ ( du／dy ) - 2yu = 2y3

I(y) = e∫- 2ydy = exp( - y2 )

u = exp( y2 )[∫y3exp( - y2 )dy + c ]

Let τ = y2 → dτ = 2ydy

→∫y3exp( - y2 )dy = ( 1／2 )∫τ e - τ dτ

= ( 1／2 )( - τ e - τ - e - τ ) + c

= ( 1／2 )[ - y2exp( - y2 ) - exp( - y2 ) ] + c

→ u = exp( y2 )[∫y3exp( - y2 )dy + c ]

= exp( y2 ){ ( 1／2 )[ - y2exp( - y2 ) - exp( - y2 ) ] + c }

= c exp( y2 ) - ( 1／2 )( y2 + 1 )

→ u = c exp( y2 ) - ( 1／2 )( y2 + 1 )

→ x2 = c exp( y2 ) - ( 1／2 )( y2 + 1 ) #