What is the solution of the differential equation \[y - x\dfrac{{dy}}{{dx}} = a\left( {{y^2} + \dfrac{{dy}}{{dx}}} \right)\]?
A. \[\left( {x + a} \right)\left( {x + ay} \right) = cy\]
B. \[\left( {x + a} \right)\left( {1 - ay} \right) = cy\]
C. \[\left( {x + a} \right)\left( {1 - ay} \right) = c\]
D. None of these

Hint: First we will find \[\dfrac{{dy}}{{dx}}\] in terms of x, y, a. Then separate all variables and take integration on both sides of the differential equation. Then integrating both sides of the differential equation and apply the logarithm formula to simplify the solution.

Formula Used: \[\int {\dfrac{1}{x}dx} = \log x + c\]
Product formula of logarithm:
\[\log ab = \log a + \log b\]
Quotient formula of logarithm:
\[\log \dfrac{a}{b} = \log a - \log b\]

Complete step by step solution: Given differential equation is:
\[y - x\dfrac{{dy}}{{dx}} = a\left( {{y^2} + \dfrac{{dy}}{{dx}}} \right)\]
\[ \Rightarrow y - x\dfrac{{dy}}{{dx}} = a{y^2} + a\dfrac{{dy}}{{dx}}\]
Add both sides by \[x\dfrac{{dy}}{{dx}}\]
\[ \Rightarrow y - x\dfrac{{dy}}{{dx}} + x\dfrac{{dy}}{{dx}} = a{y^2} + a\dfrac{{dy}}{{dx}} + x\dfrac{{dy}}{{dx}}\]
Calculating the value of \[\dfrac{{dy}}{{dx}}\]
\[ \Rightarrow y = a{y^2} + \left( {a + x} \right)\dfrac{{dy}}{{dx}}\]
\[ \Rightarrow y - a{y^2} = \left( {a + x} \right)\dfrac{{dy}}{{dx}}\]
Now we will separate the variables of the differential equation:
\[ \Rightarrow \dfrac{{dx}}{{\left( {a + x} \right)}} = \dfrac{{dy}}{{y - a{y^2}}}\]
\[ \Rightarrow \dfrac{{dx}}{{\left( {a + x} \right)}} = \dfrac{{dy}}{{y\left( {1 - ay} \right)}}\]
We can rewrite \[\dfrac{1}{{y\left( {1 - ay} \right)}}\] as sum of two terms.
\[\dfrac{1}{{y\left( {1 - ay} \right)}} = \dfrac{1}{y} + \dfrac{a}{{1 - ay}}\]
Substitute \[\dfrac{1}{{y\left( {1 - ay} \right)}} = \dfrac{1}{y} + \dfrac{a}{{1 - ay}}\] in \[\dfrac{{dx}}{{\left( {a + x} \right)}} = \dfrac{{dy}}{{y\left( {1 - ay} \right)}}\]
\[ \Rightarrow \dfrac{{dx}}{{\left( {a + x} \right)}} = \left( {\dfrac{1}{y} + \dfrac{a}{{1 - ay}}} \right)dy\]
\[ \Rightarrow \dfrac{{dx}}{{\left( {a + x} \right)}} = \dfrac{1}{y}dy + \dfrac{a}{{1 - ay}}dy\]
Taking integration on both sides:
\[ \Rightarrow \int {\dfrac{{dx}}{{\left( {a + x} \right)}}} = \int {\dfrac{1}{y}dy} + \int {\dfrac{a}{{1 - ay}}dy} \]
Assume that \[{I_1} = \int {\dfrac{a}{{1 - ay}}dy} \]
Let \[1 - ay = z\]
Differentiate both sides:
\[ - ady = dz\]
Substitute \[ - ady = dz\] and \[1 - ay = z\] in \[{I_1}\]
\[{I_1} = - \int {\dfrac{1}{z}dz} \]
Applying the formula \[\int {\dfrac{1}{x}dx} = \log x + c\]
\[{I_1} = - \log z + {c_1}\]
Substitute the value of z:
\[{I_1} = - \log \left( {1 - ay} \right) + {c_1}\]
By applying \[\int {\dfrac{1}{x}dx} = \log x + c\],we get \[\int {\dfrac{{dx}}{{\left( {a + x} \right)}}} = \log x + {c_2}\]
By applying \[\int {\dfrac{1}{x}dx} = \log x + c\],we get \[\int {\dfrac{{dy}}{y}} = \log y + {c_3}\]
Substitute the values of \[\int {\dfrac{{dx}}{{\left( {a + x} \right)}}} ,\,\int {\dfrac{1}{y}dy} ,\,\int {\dfrac{a}{{1 - ay}}dy} \] in equation (i)
\[ \Rightarrow \log \left( {a + x} \right) = \log y - \log \left( {1 - ay} \right) + \log c\]
Applying quotient formula of logarithm:
\[ \Rightarrow \log \left( {a + x} \right) = \log \dfrac{y}{{\left( {1 - ay} \right)}} + \log c\]
Now applying the product formula:
\[ \Rightarrow \log \left( {a + x} \right) = \log \dfrac{{cy}}{{\left( {1 - ay} \right)}}\]
Simplify the above equation using antilogarithm formula:
\[ \Rightarrow \left( {a + x} \right) = \dfrac{{cy}}{{\left( {1 - ay} \right)}}\]
Multiply both sides by \[\left( {1 - ay} \right)\]:
\[ \Rightarrow \left( {a + x} \right)\left( {1 - ay} \right) = cy\]

Option ‘B’ is correct

Note: Students often do a mistake when they integrate \[\int {\dfrac{a}{{1 - ay}}dy} \]. When they used substitution method, they often miss to put negative sign during substitution. They wrote \[\int {\dfrac{a}{{1 - ay}}dy} = \int {\dfrac{1}{z}dz} \] which is incorrect. The correct one is \[\int {\dfrac{a}{{1 - ay}}dy} = - \int {\dfrac{1}{z}dz} \].