# Intergrating a Differential Equation using the intergrating factor method

## 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 :)

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

Which makes your integrating factor $$x^\frac 1 2$$

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

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

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

Thanks :)

Last edited:
$$yx^{\frac{1}{2}}= \int{x^{\frac{1}{2}}x^-{\frac{1}{2}}}dx$$

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?

$$x^{\frac{1}{2}}x^{-\frac{1}{2}}$$

is what i mean

thanks :)

$$x^{\frac{1}{2}}x^{-\frac{1}{2}}$$

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, $$-\int x^\frac 1 2x^\frac 1 2dx$$.