Question

Solve the following initial value problem dy/dt = 1-y^2 y(0)=1

Answer 1

@Loser66 @surjithayer @ganeshie8

Answer 2

(In case you didn't notice, I'm not too sure how to solve the integral 1/(1-y^2) )

Answer 3

(If I were to use separation of variables..)

Answer 4

(And I can't do linear case either cause y^2 is obviously not linear)..

Answer 5

partial fractions

Answer 6

1/(1-y)(y+1) = A/(1-y) + B/(1+y)
find A,B

Answer 7

\[\frac{ dy }{1-y^2 }=dt,
\frac{ 1 }{1-y^2 }=\frac{ 1 }{ \left( 1-y \right)\left( 1+y \right) }=\frac{ 1 }{ \left( 1-y \right)\left( 1+1 \right) }+\frac{ 1 }{\left(1-(-1) \right)\left( 1+y \right) }\]
\[=\frac{ 1 }{2\left( 1-y \right) }+\frac{ 1 }{2\left( 1+y \right) }\]

\[\frac{ 1 }{ 2 }\left( \frac{ 1 }{ 1-y } +\frac{ 1 }{ 1+y }\right)dy=dt\]
integrating both sides
\[-\ln \left( 1-y \right)+\ln \left( 1+y \right)=2t+c\]
\[\ln \frac{ 1+y }{ 1-y }=2t+c\]
when t=0,y=1
here is problem for 1-1=0

Answer 8

\[1+y=\left( 1-y \right) e ^{2t+c}=C \left( 1-y \right)e ^{2t}\]
C can not be found by these values.