Question

Solve the Differential Equation
\[2(1-x^2)\frac{\text dy}{\text dx}-(1+x)y=(1-x^2)^{1/2}\]

Answer 1

\[2(1-x^2)\frac{\text dy}{\text dx}-(1+x)y=(1-x^2)^{1/2}\]

Answer 2

is it a Bernoulli equation/?

Answer 3

Get dy/dx by itself. It does look like a Bernoulli.

Answer 4

\[\frac{dy}{dx}-\frac{(1+x)}{2(1-x^2)}y=\frac{(1-x^2)^{1/2}}{2(1-x^2)}\]
\[\frac{dy}{dx}-\frac{(1+x)}{2(1-x^2)}y=\frac{(1-x^2)^{-1/2}}{2}\]

Answer 5

A Bernoulli Equation has the form\[y'+p(x)y=q(x)y^n\]

Answer 6

so
\[p(x)=-\frac{(1+x)}{2(1-x^2)}\]\[q(x)=\frac{(1-x^2)^{-1/2}}{2}\]\[n=0\]

Answer 7

is this right?

Answer 8

yes

Answer 9

so i substitute
\[v=y^{1-n}=y\]

Answer 10

Find the integrating factor now

Answer 11

\[I.F=e^{\int\limits_{}^{} P(x)}\]

Answer 12

integrating factor is
\[R(x)=e^{\int p(x)\text dx}=e^{\int-\frac{(1+x)}{2(1-x^2)}\text dx}\]
\[=e^{-\frac 12\int\frac{(1+x)}{(1-x^2)}\text dx}\]

Answer 13

Integration is real easy one

Answer 14

should i use a substitution now
easy? i can not see how it is easy

Answer 15

or should break the integral into two parts?

Answer 16

\[\frac{1+x}{1-x^2}=\frac{1+x}{(1+x)(1-x)}\]

Answer 17

ah . clever

Answer 18

integrating factor is
\[=e^{-\frac {1}{2}\ln(1-x)}=(1-x)^{-1/2}\]

Answer 19

right?

Answer 20

see it doesn't get any better

Answer 21

alrighty then ,
i should be able to solve from here i think.
(my computers is about to run outta batteries)
if i can get this question done , ill be back tomorrow,
but i think ill be right from here)
Thankyou @NotSObright

Answer 22

\[2(1-x^2)\frac{\text dy}{\text dx}-(1+x)y=(1-x^2)^{1/2}\]
\[\frac{\text dy}{\text dx}-\frac{(1+x)}{2(1-x^2)}y=\frac{(1-x^2)^{1/2}}{2(1-x^2)}\]
\[\frac{\text dy}{\text dx}-\frac{(1+x)}{2(1-x^2)}y=\frac{1}{2(1-x^2)^{1/2}}\]
\[\frac{\text dy}{\text dx}-\frac 12\frac{(1+x)}{(1+x)(1-x))}y=\frac{1}{2(1-x^2)^{1/2}}\]
\[\frac{\text dy}{\text dx}-\frac 12\frac{1}{(1-x)}y=\frac{1}{2(1-x^2)^{1/2}}\]
integrating factor
\[R(x)=e^{-\frac 12\int\frac{1}{(1-x)}\text dx}=e^{\frac 12 \ln(1-x)\text dx}=(1-x)^{1/2}\]
\[\frac{\text d }{\text dx}\left({y(1-x)^{1/2}}\right)=\frac 12\frac{(1-x)^{1/2}}{(1-x^2)^{1/2}}\]
\[\frac{\text d }{\text dx}\left({y(1-x)^{1/2}}\right)=\frac 12\left(\frac{1-x}{(1-x)(1+x)}\right)^{1/2}\]
\[y(1-x)^{1/2}=\frac 12\int\frac 1{\sqrt{(1+x)}}\text dx\]
\[\text{let }1+x=w\]
\[{\text dx}={\text dw}\]
\[y\sqrt{1-x}=\frac 12\int w^{-1/2}\text dw\]
\[y\sqrt{1-x}=\frac 12 \frac{w^{1/2}}{1/2}+c\]
\[y\sqrt{1-x}=w^{1/2}+c\]
\[y\sqrt{1-x}=\sqrt{1+x}+c\]