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

[kscplay]

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.

 

[ehild]

Hi kscplay,

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

ehild

 

[Mark44]

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.

 

[ehild]

I meant (x+√2sin[√2xy])²=-2+2sin²[√2xy]).

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

 

[jkristia]

this might be a silly question, but wouldn't you first isolate y?

 

[ehild]

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

 

[ehild]

this might be a silly question, but wouldn't you first isolate y?

If we isolate y we get y=\frac{1}{\sqrt{2}x}\arcsin(-\frac{x^2+2}{2\sqrt{2}x}) 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+\sqrt{2}\sin(\sqrt{2}xy)^2=-2+2sin^2(\sqrt{2}xy) shows that the left hand side can not be negative, the right-hand side can not be positive so both have to be zero.

x+\sqrt{2}\sin(\sqrt{2}xy)=0
\sin^2(\sqrt{2}xy)=1

two equations, two unknowns...

ehild

 

[Mentallic]

If we isolate y we get y=\frac{1}{\sqrt{2}x}\arcsin(-\frac{x^2+2}{2\sqrt{2}x}) 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.

x+\sqrt{2}\sin^2(\sqrt{2}xy)=0
\sin^2(\sqrt{2}xy)=1

two equations, two unknowns...

ehild

That's genius! I'm glad I didn't miss out on reading this

 

[ehild]

There was a typo , I corrected it in post #7. The square should not be there in my first equation. Correctly it is x+\sqrt{2}\sin(\sqrt{2}xy)=0
ehild

 

[kscplay]

That's a smart idea. Thanks ehild :)

 

[ehild]

You are welcome, kscplay.

The other way, jkristia suggested would work too.

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

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

The function

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

ehild