`3(y-4x^2)dx + xdy = 0` Solve the first-order differential equation by any appropriate method

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Given` 3(y-4x^2)dx + xdy = 0`

=>` 3y - 12x^2 +xdy/dx=0`

=>` ( 3y - 12x^2)/x +dy/dx=0`

=>` 3y/x - 12x +dy/dx=0`

=> `y'+(3/x)y=12x`

when the first order linear ordinary differential equation has the form of

`y'+p(x)y=q(x)`

then the general solution is ,

`y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/ e^(int p(x) dx) `

so,

` y'+(3/x)y=12x--------(1)`

`y'+p(x)y=q(x)---------(2)`

on comparing both we get,

`p(x) = (3/x) and q(x)=12x`

so on solving with the above general solution we get:

y(x)=`((int e^(int p(x) dx) *q(x)) dx +c)/ e^(int p(x) dx)`

=`(int e^(int (3/x) dx) *((12x)) dx +c)/e^(int (3/x) dx)`

first we shall solve

`e^(int (3/x) dx)=e^(ln(x^3))=x^3`     

so proceeding further, we get

y(x) =`(int e^(int (3/x) dx) *((12x)) dx +c)/e^(int (3/x) dx)`

=`(int x^3 *((12x)) dx +c)/x^3`

=`(int 12x^4 dx +c)/x^3`

 `=(12x^5/5 +c ) /x^3`

so `y=(12x^5/5 +c )/x^3`

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