`(dy)/(dx)-3/xy=7`

Here, `P(x)=-3/x` and `Q(x) = 7`.

Now for the integrating factor:

`"IF"=e^(intPdx)` `=e^(int-3/xdx` `=e^(-3 ln\ x)` `=e^(ln\ x^-3)` `=x^-3`

[If you are not sure what just happened, check out Integration: Basic Logarithm Form and Logarithms to Base *e*.]

We need to apply the following formula: `ye^(intPdx)=int(Qe^(intPdx))dx`

For the left hand side of the formula, we have

`ye^(intPdx) = yx^-3`

For the right hand of the formula, *Q* = 7 and the IF = *x*^{-3}, so:

`Qe^(intPdx)=7x^-3`

Applying the outer integral:

`int(Qe^(intPdx))dx=int7x^-3dx =` `-7/2x^-2+K`

Now, applying the whole formula, `ye^(intPdx)=intQe^(intPdx)dx`, we have

`yx^-3=-7/2x^-2+K`

Multiplying throughout by *x*^{3} gives the general solution for `y`.:

`y=-7/2x+Kx^3`

Is it correct? Differentiate this answer to make sure it produces the differential equation in the question.

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