Intergrating a Differential Equation using the intergrating factor method

[thomas49th]

Homework Statement

\frac{dy}{dx} + \frac{y}{2x} = -x^{\frac{1}{2}}

Homework Equations

yT =\int{QT}dx + C
where T is the intergrating factor

T = e^{\int{P}dx
and P is the co-efficient of y from the differential equations

The Attempt at a Solution

well to find T we need to do:

e^{\int{\frac{1}{2x}}}dx
e^{\frac{1}{2}\int{\frac{1}{x}}}dx
e^{\frac{1}{2}ln|x|}
= x^{\frac{1}{2}}

so using yT =\int{QT}dx + C

you get

yx^{\frac{1}{2}}= \int{x^{\frac{1}{2}x^-{\frac{1}{2}}}dx
yx^{\frac{1}{2}}= \int{1}dx
yx^{\frac{1}{2}}= x + c

the answer in the back of the book says yx^{\frac{1}{2}} = \frac{1}{2}x^{2}
Where have I gone wrong?

Thanks :)

 

[rocomath]

\frac{dy}{dx}-\left(-\frac{1}{2x}\right)y=-x^{\frac{1}{2}}

Which makes your integrating factor x^\frac 1 2

yx^\frac 1 2=-\int x^\frac 1 2 x^\frac 1 2dx=-\int xdx

Watch your inputting.

 

[thomas49th]

<br /> yx^{\frac{1}{2}}= \int{x^{\frac{1}{2}}x^-{\frac{1}{2}}}dx<br />

woops. above is what is ment to say
x^{\frac{1}{2}}x^{-\frac{1}{2}}
simplifies to 1 right?

Thanks :)

 

[rocomath]

<br /> yx^{\frac{1}{2}}= \int{x^{\frac{1}{2}}x^-{\frac{1}{2}}}dx<br />

woops. above is what is ment to say
x^{\frac{1}{2}}x^-{\frac{1}{2}}
simplifies to 1 right?

Thanks :)

No! Is the negative with x^1/2 or with just the 1/2?

 

[thomas49th]

<br /> x^{\frac{1}{2}}x^{-\frac{1}{2}}<br />

is what i mean

thanks :)

 

[rocomath]

<br /> x^{\frac{1}{2}}x^{-\frac{1}{2}}<br />

is what i mean

thanks :)

I knew what you meant, but why would the negative be with the 1/2 making it x^{-\frac 1 2}?

Isn't the right side -\int x^\frac 1 2dx then multiplied by your integrating factor, -\int x^\frac 1 2x^\frac 1 2dx.

 

[thomas49th]

your a genius :)