# Thread: Differential Equations Intergating Factors

1. ## Differential Equations Intergating Factors

Hi

Having trouble with the following problem:

Solve: $\displaystyle \frac{dy}{dx} - \frac{4y}{x} = x^4$
$\displaystyle \frac{dy}{dx} - 4y = x^5$

$\displaystyle e^{\int p(x) dx}$

$\displaystyle e^{\int-4y dx}$

$\displaystyle e^{-2y^2}$

$\displaystyle e^{-2y^2}(\frac{dy}{dx}-4y)=e^{-2y^2}x^5$

$\displaystyle e^{-2y^2}y=e^{-2y^2}x^5$

$\displaystyle e^{-2y^2}y=\int e^{-2y^2}x^5$

$\displaystyle udv = uv - \int vdu$

$\displaystyle u=e^{-2y^2}$ $\displaystyle du=-4ye^{-2y^2}$

$\displaystyle dv=x^5$ $\displaystyle v=\frac{x^6}{6}$

$\displaystyle e^{-2y^2}x^5 = \frac{e^{-2y^2}x^6}{6} - \int{\frac{x^6}{6} * -4ye^{-2y^2}}$

$\displaystyle e^{-2y^2}x^5 = \frac{e^{-2y^2}x^6}{6} - \int{\frac{x^6}{6} * -4ye^{-2y^2}}$

i continued doing the integration and i got nowhere, where is my mistake??

P.S

2. Originally Posted by Paymemoney
Hi

Having trouble with the following problem:

Solve: $\displaystyle \frac{dy}{dx} - \frac{4y}{x} = x^4$
$\displaystyle \frac{dy}{dx} - 4y = x^5$

$\displaystyle e^{\int p(x) dx}$

$\displaystyle e^{\int-4y dx}$

$\displaystyle e^{-2y^2}$

$\displaystyle e^{-2y^2}(\frac{dy}{dx}-4y)=e^{-2y^2}x^5$

$\displaystyle e^{-2y^2}y=e^{-2y^2}x^5$

$\displaystyle e^{-2y^2}y=\int e^{-2y^2}x^5$

$\displaystyle udv = uv - \int vdu$

$\displaystyle u=e^{-2y^2}$ $\displaystyle du=-4ye^{-2y^2}$

$\displaystyle dv=x^5$ $\displaystyle v=\frac{x^6}{6}$

$\displaystyle e^{-2y^2}x^5 = \frac{e^{-2y^2}x^6}{6} - \int{\frac{x^6}{6} * -4ye^{-2y^2}}$

$\displaystyle e^{-2y^2}x^5 = \frac{e^{-2y^2}x^6}{6} - \int{\frac{x^6}{6} * -4ye^{-2y^2}}$

i continued doing the integration and i got nowhere, where is my mistake??

P.S
p(x) = -4/x NOT -4y.

3. Also, you can't multiply both sides by $\displaystyle x$ without an $\displaystyle x$ ending up attached to $\displaystyle \displaystyle \frac{dy}{dx}$.

4. ok i have tried it again my answer does not match the book's answers.

$\displaystyle \frac{dy}{dx} - \frac{4y}{x} = x^4$

$\displaystyle e^\int{ln(x^4)} = x^4$

$\displaystyle x^4(\frac{dy}{dx} - \frac{4y}{x}) = x^8$

$\displaystyle (\frac{d(x^4y)}{dx} = x^8$

$\displaystyle x^4y = \int x^8$

$\displaystyle x^4y = \frac{x^9}{9} + C$

$\displaystyle y = \frac{x^5}{9} + \frac{C}{x^4}$

5. Originally Posted by Paymemoney
ok i have tried it again my answer does not match the book's answers.

$\displaystyle \frac{dy}{dx} - \frac{4y}{x} = x^4$

$\displaystyle e^{\int \ln(x^4) \, dx} = x^4$ Mr F says: I have made some formatting and notational changes to this line without changing its content. My request to you is this: Please explain where it has come from, noting that, as has previously been pointed out to you, p(x) = -4/x.

$\displaystyle x^4(\frac{dy}{dx} - \frac{4y}{x}) = x^8$

$\displaystyle (\frac{d(x^4y)}{dx} = x^8$

$\displaystyle x^4y = \int x^8$

$\displaystyle x^4y = \frac{x^9}{9} + C$

$\displaystyle y = \frac{x^5}{9} + \frac{C}{x^4}$
..

6. yep i know where i mistake, was i got the right answer now

7. As already pointed out you need

$\displaystyle \displaystyle e^{\int \frac{-4}{x}~dx} = e^{-4\ln x} = e^{\ln x^{-4}} = x^{-4}$

Now multiply this guy through the entire equation.