y'=ycosx/(1+y^2), y(0)=1

Here is my method.
y=integral of ycosx/(1+y^2) dx
1+y^2=sinx+C
Why is this wrong?

Question

y'=ycosx/(1+y^2), y(0)=1

Here is my method.
y=integral of ycosx/(1+y^2) dx
1+y^2=sinx+C
Why is this wrong?

ganeshie8 · Answer

Look at ur right hand side  :

y'=ycosx/(1+y^2)

ganeshie8 · Answer

when u dont knw "y", how can u take integral of "y"  with.respect.to. ANOTHER VARIABLE x ?

ganeshie8 · Answer

your task is to find "y", 
so first step is to write y'  as  dy/dx

ganeshie8 · Answer

$\large    y'=\frac{y\cos x}{1+y^2} $

$\large    \frac{dy}{dx}=\frac{y\cos x}{1+y^2} $

ganeshie8 · Answer

separate variables by cross multiplying

ganeshie8 · Answer

$\large \frac{1+y^2}{y}~dy = \cos x  ~dx$

$\large \frac{1}{y} +  y ~dy = \cos x  ~dx$

ganeshie8 · Answer

Integrate both sides now

ganeshie8 · Answer

$\large \int    \frac{1}{y} +  y ~dy = \int   \cos x  ~dx$


$\large \ln y +  \frac{y^2}{2} = \sin x + C$

ganeshie8 · Answer

see if that makes more or less sense..  .