# Thread: 2 first order d.e problems

1. ## 2 first order d.e problems

Urg, im really struggling with D.E...

1.
nvm i found what i was doing wrong in this one

2.
$\displaystyle \frac{dy(sin x)}{dx} + ysecx = cos^2x$
I get an answer with a bunch of e's in it :S.
$\displaystyle y = cosx + kcotx$

Any help for either questions would be much appriciated

2. Originally Posted by AshleyT
Urg, im really struggling with D.E...

1.
$\displaystyle \frac{dy}{dx} \frac{-3y}{x} = x$
I get:
Where the integrating factor is -x^3
Thus:
$\displaystyle \frac{d(-yx^3)}{dx} = -x^4$
And i get for an answer:
$\displaystyle y = \frac{x^2}{5} - \frac{c}{x^3}$
$\displaystyle y = kx^3 - x^2$

I don't get what im doing wrong :S

And this one i get a completely wrong answer
2.
$\displaystyle \frac{dy(sin x)}{dx} + ysecx = cos^2x$
I get an answer with a bunch of e's in it :S.
$\displaystyle y = cosx + kcotx$

Any help for either questions would be much appriciated
For the first, you are correct, teh integrating factor is $\displaystyle x^{-3}$ but there's a mistake in what follows

$\displaystyle \frac{1}{x^3} \frac{dy}{dx} - 3 \frac{1}{x^4} = \frac{1}{x^2}$ so $\displaystyle \frac{d}{dx}\left( \frac{y}{x^3}\right) = \frac{1}{x^2}$.

Integrating gives

$\displaystyle \frac{y}{x^3} = - \frac{1}{x} + k$ which leads to the answer.

For the second problem, first put in standard form

$\displaystyle \frac{dy}{dx} + \frac{y}{\sin x \cos x} = \frac{\cos^2x}{\sin x}$

the integrating factor is $\displaystyle \tan x$ so

$\displaystyle \frac{d}{dx} \left( \tan x \cdot y \right) = \cos x$ which you can integrate.

3. Originally Posted by danny arrigo
For the first, you are correct, teh integrating factor is $\displaystyle x^{-3}$ but there's a mistake in what follows

$\displaystyle \frac{1}{x^3} \frac{dy}{dx} - 3 \frac{1}{x^4} = \frac{1}{x^2}$ so $\displaystyle \frac{d}{dx}\left( \frac{y}{x^3}\right) = \frac{1}{x^2}$.

Integrating gives

$\displaystyle \frac{y}{x^3} = - \frac{1}{x} + k$ which leads to the answer.

For the second problem, first put in standard form

$\displaystyle \frac{dy}{dx} + \frac{y}{\sin x \cos x} = \frac{\cos^2x}{\sin x}$

the integrating factor is $\displaystyle \tan x$ so

$\displaystyle \frac{d}{dx} \left( \tan x \cdot y \right) = \cos x$ which you can integrate.
However how do you achieve tan x?
You integrate cosecxsecx right? I thought this was sec x?
Then to e ^ sec x

This must have been what i was doing wrong :S

4. Originally Posted by AshleyT
However how do you achieve tan x?
You integrate cosecxsecx right? I thought this was sec x?
Then to e ^ sec x

This must have been what i was doing wrong :S
For your integrating factor $\displaystyle \int \sec x \csc x\, dx = \ln \tan x$
just to note $\displaystyle \int \sec x \tan x\, dx = \sec x$

5. Originally Posted by danny arrigo
For your integrating factor $\displaystyle \int \sec x \csc x\, dx = \ln \tan x$
just to note $\displaystyle \int \sec x \tan x\, dx = \sec x$
oic.
Thank-you