Question

What is the solution of the equation \[\left( {1 - {x^2}} \right)dy + xydx = x{y^2}dx\]?
A. \[{\left( {y - 1} \right)^2}\left( {1 - {x^2}} \right) = 0\]
B. \[{\left( {y - 1} \right)^2}\left( {1 - {x^2}} \right) = {c^2}{y^2}\]
C. \[{\left( {y - 1} \right)^2}\left( {1 + {x^2}} \right) = {c^2}{y^2}\]
D. None of these

Answer 1

Hint: First we separate the variables of the given differential equation. Then simplify the equation after that we use the substitution method to apply the integration formula. Then solve the differential equation to get the result.

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

Complete step by step solution:
Given differential equation is
\[\left( {1 - {x^2}} \right)dy + xydx = x{y^2}dx\]
Separate the variables of the above equation:
\[ \Rightarrow \left( {1 - {x^2}} \right)dy = x{y^2}dx - xydx\]
\[ \Rightarrow \left( {1 - {x^2}} \right)dy = xy\left( {y - 1} \right)dx\]
Divide both sides by \[\left( {1 - {x^2}} \right)y\left( {y - 1} \right)\]
\[ \Rightarrow \dfrac{{\left( {1 - {x^2}} \right)}}{{\left( {1 - {x^2}} \right)y\left( {y - 1} \right)}}dy = \dfrac{{xy\left( {y - 1} \right)}}{{\left( {1 - {x^2}} \right)y\left( {y - 1} \right)}}dx\]
Cancel out common term from denominator and numerator
\[ \Rightarrow \dfrac{1}{{y\left( {y - 1} \right)}}dy = \dfrac{x}{{\left( {1 - {x^2}} \right)}}dx\] …..(i)
We know that, \[\dfrac{1}{{y\left( {y - 1} \right)}} = \dfrac{1}{{y - 1}} - \dfrac{1}{y}\]
Substitute \[\dfrac{1}{{y\left( {y - 1} \right)}} = \dfrac{1}{{y - 1}} - \dfrac{1}{y}\] in equation (i)
\[ \Rightarrow \left( {\dfrac{1}{{y - 1}} - \dfrac{1}{y}} \right)dy = \dfrac{x}{{\left( {1 - {x^2}} \right)}}dx\] …..(ii)
Assume that, \[\left( {1 - {x^2}} \right) = z\]
Differentiate the above equation:
\[ - 2xdx = dz\]
Substitute \[\left( {1 - {x^2}} \right) = z\] and \[2xdx = dz\] in equation (ii)
\[ \Rightarrow \left( {\dfrac{1}{{y - 1}} - \dfrac{1}{y}} \right)dy = - \dfrac{1}{{2z}}dz\]
Apply antiderivative in the above equation
\[ \Rightarrow \int {\dfrac{1}{{y - 1}}dy} - \int {\dfrac{1}{y}dy} = - \int {\dfrac{1}{{2z}}dz} \]
\[ \Rightarrow \log \left| {y - 1} \right| - \log \left| y \right| = - \dfrac{1}{2}\log \left| z \right| + \log c\]
Multiply both sides by 2
\[ \Rightarrow 2\log \left| {y - 1} \right| - 2\log \left| y \right| = - \log \left| z \right| + 2\log c\]
Applying logarithm power rule
\[ \Rightarrow \log \left| {{{\left( {y - 1} \right)}^2}} \right| - \log \left| {{y^2}} \right| = - \log \left| z \right| + \log {c^2}\]
Putting \[\left( {1 - {x^2}} \right) = z\]
\[ \Rightarrow \log \left| {{{\left( {y - 1} \right)}^2}} \right| - \log \left| {{y^2}} \right| = - \log \left| {1 - {x^2}} \right| + \log {c^2}\]
Rearrange the terms:
\[ \Rightarrow \log \left| {{{\left( {y - 1} \right)}^2}} \right| + \log \left| {1 - {x^2}} \right| = \log \left| {{y^2}} \right| + \log {c^2}\]
Now applying product rule of logarithm:
\[ \Rightarrow \log \left[ {{{\left( {y - 1} \right)}^2}\left( {1 - {x^2}} \right)} \right] = \log \left| {{y^2}{c^2}} \right|\]
\[ \Rightarrow {\left( {y - 1} \right)^2}\left( {1 - {x^2}} \right) = {y^2}{c^2}\]
Hence option B is the correct option.

Note: Students often take c as integrating constant. After solving the differential equation, we get that all terms are a logarithm function. Thus, we have to take \[\log c\] as an integrating constant.