Answer to Question #320481 in Differential Equations for neela

Question

Find the linear and Bernoulli’s differential equations from the following differential equations and solve it.i)(1-x^2)dy/dx-xy=1 ii)dy/dx=xy^2-xy

Answer 1

$i:\\\left( 1-x^2 \right) y'-xy=1\\y'-\frac{x}{1-x^2}y=\frac{1}{1-x^2}-linear\,\,equation\\P\left( x \right) =-\frac{x}{1-x^2},Q\left( x \right) =\frac{1}{1-x^2}\\y=\frac{1}{\exp \left( \int{Pdx} \right)}\left( \int{Qe^{\int{Pdx}}dx}+C \right) =\\=\frac{1}{\exp \left( \int{\frac{x}{x^2-1}dx} \right)}\left( \int{\frac{1}{1-x^2}e^{\int{\frac{x}{x^2-1}dx}}dx}+C \right) =\\=\frac{1}{\exp \left( \frac{1}{2}\ln \left( 1-x^2 \right) \right)}\left( \int{\frac{1}{1-x^2}e^{\frac{1}{2}\ln \left( 1-x^2 \right)}dx}+C \right) =\\=\frac{1}{\sqrt{1-x^2}}\left( \int{\frac{1}{\sqrt{1-x^2}}dx}+C \right) =\frac{1}{\sqrt{1-x^2}}\left( \mathrm{a}\sin x+C \right) \\ii:\\y'+xy=xy^2-Bernoulli\,\,equation\\y=0 -\,\,trivial\\P\left( x \right) =x,Q\left( x \right) =x,n=2\\u=y^{-1}\\y=\frac{1}{u}\\y'=-\frac{1}{u^2}u'\\-\frac{1}{u^2}u'+\frac{x}{u}=\frac{x}{u^2}\\u'-xu=-x\\u'=x\left( u-1 \right) \\\frac{d\left( u-1 \right)}{u-1}=xdx\\\ln \left| u-1 \right|=\frac{x^2}{2}\\u=Ce^{\frac{x^2}{2}}+1\\y=\frac{1}{Ce^{\frac{x^2}{2}}+1}\\$