Question

SOLVE :(1-x^2)dy/dx-xy=1.?

Answer 1

I hope this is helpful for you

Answer 2

The solution of the differential equation

$(1 - {x}^{2} ) \frac{dy}{dx} - xy = 1$

is given by

$y = \frac{ {sin}^{ - 1} x + c}{ \sqrt{1 - {x}^{2} } }$

• We have,

$(1 - {x}^{2} ) \frac{dy}{dx} - xy = 1$

Now, dividing the above equation by

$(1 - {x}^{2} )$

we get,

$\frac{dy}{dx} - \frac{x}{1 - {x}^{2}}y = \frac{1}{1 - {x}^{2} }$

• Now, it is in the form of

$\frac{dy}{dx} + p(x)y \: = q(x)$

where

$p(x) = \frac{ - x}{1 - {x}^{2} } \: and \: q(x) = \frac{1}{1 - {x}^{2} }$

• Now, for solving this equation we will find integrating factor

$i.f \: = {e}^{∫p(x)dx}$

therefore,

$i.f = {e}^{∫ \frac{ - x}{1 - {x}^{2} }dx }$

Let 1 - x² = t - (1)

-xdx = dt/2

$= {e}^{ ∫ \frac{1}{2t}dt}$

$= {e}^{ \frac{1}{2} log(t) }$

$= {e}^{ log( \sqrt{t} ) }$

Now putting the value of 't' from (1) we get,

$i.f = {e}^{ log( \sqrt{1 - {x}^{2} } ) }$

$i.f = \sqrt{1 - {x}^{2} }$

• Now the solution of the differential equation is given by,

$y(i.f) \: = ∫q(x)(i.f)dx$

therefore,

$y( \sqrt{1 - {x}^{2} }) = ∫ \frac{1}{1 - {x}^{2} } ( \sqrt{1 - {x}^{2} } )dx$

$y \sqrt{1 - {x}^{2} } = ∫ \frac{1}{ \sqrt{1 - {x}^{2} } } dx$

$y \sqrt{1 - {x}^{2} } = {sin}^{ - 1} x + c$

hence the solution is,

$y = \frac{ {sin}^{ - 1} x + c}{ \sqrt{1 - {x}^{2} } }$