A linear differential equation of the first order can be either of the following forms 

(i) \(\frac{{dy}}{{dx}} + Py = Q,\) where \(P\) and \(Q\) are functions of \(x\) or constants

(ii) \(\frac{{dx}}{{dy}} + Px = Q,\) where \(P\) and \(Q\) are functions of \(y\) or constants.

Consider the differential equation (i) i.e. \(\frac{{dy}}{{dx}} + Py = Q\)

The expression \({e^{\int {Pdx} }} = R\) is called the Integrating Factor of the equation.

Multiply both sides by \({e^{\int {Pdx} }}\), so that we get,

\({e^{\int {Pdx} }}\left[ {\cfrac{{dy}}{{dx}} + Py} \right] = Q{e^{\int {Pdx} }}\)

The left hand side is the differential coefficient of \(y{e^{\int {Pdx} }}\)or \(yR\)

Therefore, the solution can also be written as 

\[yR = \int {Q.Rdx + C} \](For the second differential equation \(\frac{{dx}}{{dy}} + Px = Q,\)the integrating factor, I.F. =\({e^{\int {P\,dy} }}\)and the general solution is \(xR = \int {Q.R\;dy + C} \))

Example 1: Solve \(y\log y\frac{{dx}}{{dy}} + x - \log y = 0\).

Solution:  \(\frac{{dx}}{{dy}} + x\left( {\frac{1}{{y\log y}}} \right) = \frac{1}{y}\)

I.F., \(R = {e^{\int {\frac{1}{{y\log y}}dy} }} = {e^{\log (\log \,y)}} = \log y\)

Now the solution is: \(x.\log y = \int {\frac{1}{y}.\log ydy} \)  

\( \Rightarrow x\log y = \frac{{{{(\log y)}^2}}}{2} + C\)

Example 2: The general solution of the differential equation  \(\frac{{dy}}{{dx}} = (1 + {y^2})({e^{ - {x^2}}} - 2x{\tan ^{ - 1}}y)\) is:

(1) \({e^{{x^2}}}{\tan ^{ - 1}}y = x + c\)

(2) \({e^{ - {x^2}}}{\tan ^{ - 1}}y = x + c\)

(3) \({e^x}\tan y = {x^2} + c\)

(4) \({e^x}{\tan ^{ - 1}}y = {x^3} + c\)

Solution: Suppose \({\tan ^{ - 1}}y = z\)

 \[ \Rightarrow \frac{{dz}}{{dx}} = \frac{1}{{1 + {y^2}}}\left[ {\frac{{dy}}{{dx}}} \right]\]  

\[ \Rightarrow \frac{{dy}}{{dx}} = [1 + {y^2}\frac{{dz}}{{dx}}\]

Putting this value in the equation, we have:

\[[1 + {y^2}\frac{{dz}}{{dx}} = [1 + {y^2}{e^{ - {x^2}}} - 2xz]\] 

\[\frac{{dz}}{{dx}} + 2zx = {e^{ - {x^2}}}\]

Integrating factor \(R = {e^{\int {2xdx} }} = {e^{{x^2}}}\)

Solution of the equation is: \[zR = \int {{e^{ - {x^2}}}.{e^{{x^2}}}dx}  + c\]\[ \Rightarrow {e^{{x^2}}}{\tan ^{ - 1}}y = x + c\]