Differential Equation & Linear Algebra Help?

[JSGhost]

Homework Statement

Solve the following differential equation
8x - 2y sqrt(x^2 + 1) dy/dx = 0
subject to the initial condition: y(0) = -3.

y = ?

Homework Equations

Separable DE's
dy/dx = g(x)/f(y)

The Attempt at a Solution

8x - 2y sqrt(x^2 + 1) dy/dx = 0
8x = 2y sqrt(x^2 + 1) dy/dx
[8x/sqrt(x^2+1)]dx = 2y dy
integrate both sides

u = x^2 + 1
du = 2x dx

4 (integral) [1/sqrt(u)] du = y^2
4(2u^(1/2)) + c = y^2
8 sqrt(x^2+1) + c = y^2

Since y(0) = -3.
Substitute to find c.
8sqrt((0)^2+1) + c = (-3)^2
8 + c = 9
c = 1

y^2 = 8 sqrt(x^2+1) + 1

I get that answer but it isn't correct. I am using webwork and need help solving the equation. Thanks.

 

[tiny-tim]

Hi JSGhost!

(have a square-root; √ and try using the X² tag just above the Reply box )

Solve the following differential equation
8x - 2y sqrt(x^2 + 1) dy/dx = 0
subject to the initial condition: y(0) = -3.
…
y^2 = 8 sqrt(x^2+1) + 1

Looks ok so far …

what was your final answer, "y = … " ?

 

[JSGhost]

y^2 = 8 sqrt(x^2 + 1) + 1

to

y = sqrt( 8(x^2+1)^(1/2) + 1) ?

I've tried that already.

 

[tiny-tim]

Hi JSGhost!

(just got up :zzz: …)

y = sqrt( 8(x^2+1)^(1/2) + 1) ?

("sqrt" and "1/2" ? and what happened to that √ I gave you? )

erm …

that doesn't look negative to me! ​

 

[JSGhost]

Nothing seems to work. Maybe I am approaching the problem wrong? Maybe it's not separable...but linear first-order equation.

y = sqrt( 8sqrt(x^1 + 1) + 1) doesn't work. Those symbols don't work cause I can't copy those into the answer placement.

 

[tiny-tim]

If you put x = 0 into sqrt( 8sqrt(x^1 + 1) + 1), you get √(8√(1) + 1) = √(8+1) = 3,

and you want -3

sooo … ? ​

 

[HallsofIvy]

If $y^2= f(x)$ y is NOT necessarily equal to $\sqrt{f(x)}$! There are two possible solutions and only one can satisfy y(0)= -3.

 

[JSGhost]

Thanks. Finally got it. I needed a negative. I thought sqrt(9) would give negative and positive of 3.

y = - sqrt( 8sqrt(x^1 + 1) + 1)

 

[tiny-tim]

he he

a maths symbol only ever has its unique "principal value" …

see http://en.wikipedia.org/wiki/Principal_value"