Answer to Question #241166 in Differential Equations for Jojo

Question

(1-x^ 2 )dy/dx-x^ 2 y=(1+x)* sqrt (1- x^ 2 )

Answer 1

$\left( {1 - {x^2}} \right)\frac{{dy}}{{dx}} - {x^2}y = \left( {1 + x} \right)\sqrt {1 - {x^2}}$

$y' - \frac{{{x^2}}}{{1 - {x^2}}}y = \frac{{\left( {1 + x} \right)\sqrt {1 - {x^2}} }}{{\left( {1 - {x^2}} \right)}}$

Let

$y = uv \Rightarrow y' = u'v + uv'$

Then

$u'v + uv' - \frac{{{x^2}}}{{1 - {x^2}}}uv = \frac{{\left( {1 + x} \right)\sqrt {1 - {x^2}} }}{{\left( {1 - {x^2}} \right)}}$

$u'v + u\left( {v' - \frac{{{x^2}}}{{1 - {x^2}}}v} \right) = \frac{{\left( {1 + x} \right)\sqrt {1 - {x^2}} }}{{\left( {1 - {x^2}} \right)}}$

Let

$v' - \frac{{{x^2}}}{{1 - {x^2}}}v = 0 \Rightarrow \frac{{dv}}{{dx}} = \frac{{{x^2}}}{{1 - {x^2}}}v \Rightarrow \frac{{dv}}{v} = \frac{{{x^2}}}{{1 - {x^2}}}dx = \left( {\frac{1}{{2(x + 1)}} - \frac{1}{{2(x - 1)}} - 1} \right)dx \Rightarrow \ln v = \frac{1}{2}\ln (x + 1) - \frac{1}{2}\ln (1 - x) - x \Rightarrow \ln v = \ln \sqrt {\frac{{1 + x}}{{1 - x}}} - x \Rightarrow v = {e^{\ln \sqrt {\frac{{x + 1}}{{1 - x}}} - x}} = {e^{ - x}}\sqrt {\frac{{x + 1}}{{1 - x}}}$

Then

$u'{e^{ - x}}\sqrt {\frac{{x + 1}}{{1 - x}}} = \frac{{\left( {1 + x} \right)\sqrt {1 - {x^2}} }}{{\left( {1 - {x^2}} \right)}} \Rightarrow u' = {e^x}\frac{{(1 + x)\sqrt {1 - {x^2}} }}{{\left( {1 - {x^2}} \right)}}\sqrt {\frac{{1 - x}}{{x + 1}}} = {e^x}\frac{{\sqrt {(1 - x)(1 + x)} }}{{\left( {1 - x} \right)}}\sqrt {\frac{{1 - x}}{{x + 1}}} = {e^x} \Rightarrow u = {e^x} + C$

Then

$y = uv = \left( {{e^x} + C} \right){e^{ - x}}\sqrt {\frac{{x + 1}}{{1 - x}}} = \sqrt {\frac{{x + 1}}{{1 - x}}} + C{e^{ - x}}\sqrt {\frac{{x + 1}}{{1 - x}}}$

Answer: $y = \sqrt {\frac{{x + 1}}{{1 - x}}} + C{e^{ - x}}\sqrt {\frac{{x + 1}}{{1 - x}}}$