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:


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


Multiplying throughout by x3 gives the general solution for `y`.:


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

Please support IntMath!