Question

Give the general solution of the ODE:
y'y = sqrt(1+y^2)cos(x)

Answer 1

\[y \frac{ dy }{ dx } = \sqrt{(1+y^2} )\cos(x)\]
\[\int\limits \frac{ y }{ \sqrt{(1+y^2} ) }dy = \int\limits\cos(x)dx\]
\[\int\limits \frac{ 1 }{ \sqrt{(1+y^2} ) }d(1+y^2) = \int\limits \cos(x)dx\]

Answer 2

\[-\frac{ 1 }{ 2 } \sqrt{1+y^2} = \sin(x) + C)\]

Answer 3

would that then become\[y = 2\sin(x) - 1 +c\] ?? @abtster

Answer 4

constants and the like just change the constant

Answer 5

c + 1 = c'
etc

Answer 6

ok that makes sense, thanks! @abtster

Answer 7

nice solution

Answer 8

@abtster I dont understand your integral. how did you get dy = d(y^2+1) , i dont understand that step

Answer 9

@perl ,
\[ydy = \frac{1}{2}dy^2=\frac{1}{2}d()y^2+1)\]

O ohh, my bad
I forgot the fraction
and some more flaws

Answer 10

\[−\frac{1}{4}\sqrt{(1+y^2)}=\sin(x)+C\]

Answer 11

\[\sqrt{1+y^2}=4sin(x)+C'\]
\[1+y^2=(4sin(x)+C)^2\]
\[y^2=(4sin(x)+C)^2-1\]
\[y=\sqrt{(4sin(x)+C')^2-1}\]
or
\[y=-\sqrt{(4sin(x)+C')^2-1}\]

there might be some trigonometry identities to simplify this, but I can't find it