Question

What is the solution of the differential equation \[x \dfrac{{dy}}{{dx}} + y = {y^2}\]?
A. \[y = 1 + cxy\]
B. \[y = \log\left( {cxy} \right)\]
C. \[y + 1 = cxy\]
D. \[y = c + xy\]

Answer 1

Hint: Here, the first order differential equation is given. First, simplify the given equation. Then, integrate both sides of the equation with respect to the corresponding variables. After that, solve both integrals by using the standard integral formulas. In the end, apply the property of a logarithm to get the solution of the differential equation.

Formula used:
\[\int {\dfrac{{dx}}{x}} = \log\left( x \right)\]
The logarithmic properties:
\[\log\left( a \right) - \log\left( b \right) = \log\left( {\dfrac{a}{b}} \right)\]
\[\log\left( a \right) + \log\left( b \right) = \log\left( {ab} \right)\]

Complete step by step solution:
The given differential equation is \[x \dfrac{{dy}}{{dx}} + y = {y^2}\].

Let’s find out the solution of the given differential equation.
Simplify the given equation.
\[x \dfrac{{dy}}{{dx}} = {y^2} - y\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^2} - y}}{x}\]
\[ \Rightarrow \dfrac{{dy}}{{{y^2} - y}} = \dfrac{{dx}}{x}\]
Simplify the left-hand side of the above equation.
\[\left[ {\dfrac{1}{{y - 1}} - \dfrac{1}{y}} \right]dy = \dfrac{{dx}}{x}\]
Now integrate both sides with respect to the corresponding variables.
\[\int {\left[ {\dfrac{1}{{y - 1}} - \dfrac{1}{y}} \right]dy = } \int {\dfrac{{dx}}{x}} \]
Apply the integration rule on the left-hand side.
\[\int {\dfrac{1}{{y - 1}}} dy - \int {\dfrac{1}{y}} dy = \int {\dfrac{{dx}}{x}} \]
Solve the integrals by using the standard formula \[\int {\dfrac{{dx}}{x}} = \log\left( x \right)\].
\[\log\left( {y - 1} \right) - \log y = \log x + \log c\]
Now apply the logarithmic properties.
We get,
\[\log\left( {\dfrac{{y - 1}}{y}} \right) = \log \left( {cx} \right)\]
Equate both sides.
\[\dfrac{{y - 1}}{y} = cx\]
Solve the above equation.
\[y - 1 = cxy\]
\[ \Rightarrow y = 1 + cxy\]
Hence the correct option is A.

Note: It is necessary to use an integration constant as soon as integration is performed if we solve a first-order differential equation by the variable separable method.
If all terms of the solution are in the form of a logarithm, then the integration constant is also in the form of a logarithm.