can't solve this diff equation please help..

[Ahmedkhlil]

siny dx + sqr(2- x^2) dy

 

[Deleted member 4993]

siny dx + sqr(2- x^2) dy

This is NOT a differential equation - there is no "equal to" sign.

 

[Steven G]

This is NOT a differential equation - there is no "equal to" sign.

I thought that you like a challenge.

 

[blamocur]

This is NOT a differential equation - there is no "equal to" sign.

Since nitpicking is my second name, I'd point out that this is not an equation at all, differential or not

 

[nasi112]

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


\(\displaystyle -\int \frac{1}{\sin y} \ dy= \int \frac{1}{\sqrt{2-x^2}} \ dx\)


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

 

[Steven G]

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

I didn't know this identity. This forum is great.

 

[Deleted member 4993]

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

Where did that "= 0" come from?

\(\displaystyle -\int \frac{1}{\sin y} \ dy= \int \frac{1}{\sqrt{2-x^2}} \ dx\)

 

[Steven G]

Where did that "= 0" come from?

\(\displaystyle -\int \frac{1}{\sin y} \ dy= \int \frac{1}{\sqrt{2-x^2}} \ dx\)


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

The equation above is incorrect. Please check and modify and repost.

There are two equations above. Which one is invalid?

 

[Deleted member 4993]

There are two equations above. Which one is invalid?

The one relating '(y)' & (y/2) - without calculus.

 

[BigBeachBanana]

[math]\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)[/math]

 

[Deleted member 4993]

I didn't know this identity. This forum is great.

I suggest that we stay above use of sarcasm. Can be misconstrued.

 

[Deleted member 4993]

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

My MISTAKE ... Still at the corner......

 

[Ahmedkhlil]

This is NOT a differential equation - there is no "equal to" sign.

Yeah sorry i forgot

 

[Ahmedkhlil]

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

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


\(\displaystyle -\int \frac{1}{\sin y} \ dy= \int \frac{1}{\sqrt{2-x^2}} \ dx\)


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

Thanks for helping.... actually i made a mistake its arc siny not siny :-(

 

[BigBeachBanana]

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 :-(

[imath]\arcsin(y)\,dx + \sqrt{2- x^2}\,dy = 0[/imath] is a separable equation. Can you separate and integrate?

 

[Deleted member 4993]

Since nitpicking is my second name, I'd point out that this is not an equation at all, differential or not

I thought that was your middle name.....

 

[blamocur]

I thought that was your middle name.....

Who is nitpicking now ?

 

[Deleted member 4993]

Who is nitpicking now ?

I am NOT nitpicking - I am quibbling .......