can't solve this diff equation please help..

siny dx+2x2 dy=0\displaystyle \sin y \ dx + \sqrt{2 - x^2} \ dy = 0


1siny dy=12x2 dx\displaystyle -\int \frac{1}{\sin y} \ dy= \int \frac{1}{\sqrt{2-x^2}} \ dx


Hint: cscy+coty=cot(y2)\displaystyle \ \csc y + \cot y = \cot(\frac{y}{2})
 
siny dx+2x2 dy=0\displaystyle \sin y \ dx + \sqrt{2 - x^2} \ dy = 0
Where did that "= 0" come from?

1siny dy=12x2 dx\displaystyle -\int \frac{1}{\sin y} \ dy= \int \frac{1}{\sqrt{2-x^2}} \ dx
 
Where did that "= 0" come from?

1siny dy=12x2 dx\displaystyle -\int \frac{1}{\sin y} \ dy= \int \frac{1}{\sqrt{2-x^2}} \ dx


Hint: cscy+coty=cot(y2)\displaystyle \ \csc y + \cot y = \cot(\frac{y}{2})

The equation above is incorrect. Please check and modify and repost.
There are two equations above. Which one is invalid?
 
csc(y)+cot(y)=1sin(y)+cos(y)sin(y)=1+cos(y)sin(y)=2cos2(y2)2sin(y2)cos(y2)=cot(y2)\csc(y) + \cot(y)=\frac{1}{\sin(y)}+\frac{\cos(y)}{\sin(y)}=\frac{1+\cos(y)}{\sin(y)}=\frac{2\cos^2(\frac{y}{2})}{2\sin(\frac{y}{2})\cos(\frac{y}{2})}=\cot\left(\frac{y}{2}\right)
 
siny dx + sqr(2- x^2) dy
Thank you all for reply.. I can't believe that i made two mistake in this equation >_< .. actually it should be

arc sin (y) dx + sqr(2- x^2) dy = 0

its more difficult know its a homework proplem.... we only study line
siny dx+2x2 dy=0\displaystyle \sin y \ dx + \sqrt{2 - x^2} \ dy = 0


1siny dy=12x2 dx\displaystyle -\int \frac{1}{\sin y} \ dy= \int \frac{1}{\sqrt{2-x^2}} \ dx


Hint: cscy+coty=cot(y2)\displaystyle \ \csc y + \cot y = \cot(\frac{y}{2})
Thanks for helping.... actually i made a mistake its arc siny not siny :-(
 
Thank you all for reply.. I can't believe that i made two mistake in this equation >_< .. actually it should be

arc sin (y) dx + sqr(2- x^2) dy = 0

its more difficult know its a homework proplem.... we only study line

Thanks for helping.... actually i made a mistake its arc siny not siny :-(
arcsin(y)dx+2x2dy=0\arcsin(y)\,dx + \sqrt{2- x^2}\,dy = 0 is a separable equation. Can you separate and integrate?
 
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