Solve this equation

  • Thread starter kscplay
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  • #1
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Homework Statement



Solve for x and y:
x2 + (√8)x*sin[(√2)xy] +2 = 0

Homework Equations





The Attempt at a Solution


Other than decomposing the root 8 i don't know what else to do. any hints? thanks.
 

Answers and Replies

  • #2
ehild
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Hi kscplay,

Transform the equation into the form (x+a)2+b=0

ehild
 
  • #3
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Homework Statement



Solve for x and y:
x2 + (√8)x*sin[(√2)xy] +2 = 0
QUOTE]

Hi kscplay,

Transform the equation into the form (x+a)2+b=0
ehild, I don't see how this will lead anywhere, due to the presence of x in the sine factor.
 
  • #4
ehild
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I meant (x+√2sin[√2xy])2=-2+2sin2[√2xy]).

The left hand side can not be negative. The right hand side ?...


ehild
 
  • #5
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this might be a silly question, but wouldn't you first isolate y?
 
  • #6
ehild
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this might be a silly question, but wouldn't you first isolate y?

That could be also a way to find the solution. But I would look at the right hand side of my equation and see if it can be positive.


ehild
 
Last edited:
  • #7
ehild
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this might be a silly question, but wouldn't you first isolate y?

If we isolate y we get [tex]y=\frac{1}{\sqrt{2}x}\arcsin(-\frac{x^2+2}{2\sqrt{2}x})[/tex] The argument of arcsin has to be in the range [-1,1]. What does it mean for x?

My method of transforming the equation to [itex](x+\sqrt{2}\sin(\sqrt{2}xy)^2=-2+2sin^2(\sqrt{2}xy)[/itex] shows that the left hand side can not be negative, the right-hand side can not be positive so both have to be zero.

[tex]x+\sqrt{2}\sin(\sqrt{2}xy)=0[/tex]
[tex]\sin^2(\sqrt{2}xy)=1 [/tex]

two equations, two unknowns...

ehild
 
Last edited:
  • #8
Mentallic
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If we isolate y we get [tex]y=\frac{1}{\sqrt{2}x}\arcsin(-\frac{x^2+2}{2\sqrt{2}x})[/tex] The argument of arcsin has to be in the range [-1,1]. What does it mean for x?

My method of transforming the equation to (x+√2sin[√2xy])^2=-2+2sin^2[√2xy]) shows that the left hand side can not be negative, the right-hand side can not be positive so both have to be zero.

[tex]x+\sqrt{2}\sin^2(\sqrt{2}xy)=0[/tex]
[tex]\sin^2(\sqrt{2}xy)=1 [/tex]

two equations, two unknowns...

ehild

That's genius! I'm glad I didn't miss out on reading this :smile:
 
  • #9
ehild
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:smile:
There was a typo , I corrected it in post #7. The square should not be there in my first equation. Correctly it is [tex]x+\sqrt{2}\sin(\sqrt{2}xy)=0[/tex]
ehild
 
Last edited:
  • #10
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That's a smart idea. Thanks ehild :)
 
  • #11
ehild
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You are welcome, kscplay.:smile:

The other way, jkristia suggested would work too.

[tex]y=-\frac{1}{\sqrt{2}x}\arcsin \left( \frac{x^2+2}{2\sqrt{2}x} \right)[/tex]

The magnitude of the argument of the arcsin function can not exceed 1:
[tex] \left| \frac{x^2+2}{2 \sqrt{2} x} \right| \leq 1[/tex]

The function

[tex]
\frac{x^2+2}{2\sqrt{2}x}[/tex] has extrema at x=±√2, minimum (1) for positive x and maximum (-1) for negative x. Only these values are allowed for x.

ehild
 

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