FP3 First order differential equations

I'm having some trouble with getting the general solutions of these differential equations! They should be pretty easy, they're at the start of the misc ex

$x \frac{dy}{dx} + (x+1)y = 1$

and for -1<x<1

$(1 + x^2)\frac{dy}{dx} -xy +1 = 0$

If someone could post a method for one of them, it would be super helpful! Thank you!

I'm having some trouble with getting the general solutions of these differential equations! They should be pretty easy, they're at the start of the misc ex

$x \frac{dy}{dx} + (x+1)y = 1$

and for -1<x<1

$(1 + x^2)\frac{dy}{dx} -xy +1 = 0$

If someone could post a method for one of them, it would be super helpful! Thank you!

I'm having some trouble with getting the general solutions of these differential equations! They should be pretty easy, they're at the start of the misc ex

$x \frac{dy}{dx} + (x+1)y = 1$

and for -1<x<1

$(1 + x^2)\frac{dy}{dx} -xy +1 = 0$

If someone could post a method for one of them, it would be super helpful! Thank you!

...

\int e^{P\mathrm{d}x}Q \ \mathrm{d}x

Do you know the general form for solving 1^st ODEs?

start by dividing the coefficient of dy/dx

Yeah, I know the general method but I can't seem to get the 'product rule' bit..

$[br]x (dy/dx) + (x+1)y = 1[br](dy/dx) + (1 + 1/x)y = 1/x[br][br]I(x) = \int (1 + \frac{1}{x})dx = x + ln x[br]So multiply the original thing by xe^x[br][br]xe^x (\frac{dy}{dx} + (1 + \frac{1}{x})y = \frac{1}{x})[br]xe^x \frac{dy}{dx} + xe^x (1 + \frac{1}{x})y = e^x[br]$

Sorry the latex fracs are a bit dodgy haha, but I don't know what to do from there.. Have I made any mistakes or can I just not spot the product rule bit?

I let @aymanzayedmannan continue as he was first and I found myself tied up now.

Yeah, I know the general method but I can't seem to get the 'product rule' bit..

$[br]xe^x \frac{dy}{dx} + xe^x (1 + \frac{1}{x})y = e^x[br]$

Sorry the latex fracs are a bit dodgy haha, but I don't know what to do from there.. Have I made any mistakes or can I just not spot the product rule bit?

From there: the LHS is nothing but $\displaystyle \frac{d}{dx} \left(xe^x y\right) = e^x$

Now you can integrate both sides...

Ohh, I somehow missed that you had to use product rule on xe^x Thanks!!!

Teensy typo. :-)

Yeah, I know the general method but I can't seem to get the 'product rule' bit..

$[br]x (dy/dx) + (x+1)y = 1[br](dy/dx) + (1 + 1/x)y = 1/x[br][br]I(x) = \int (1 + \frac{1}{x})dx = x + ln x[br]So multiply the original thing by xe^x[br][br]xe^x (\frac{dy}{dx} + (1 + \frac{1}{x})y = \frac{1}{x})[br]xe^x \frac{dy}{dx} + xe^x (1 + \frac{1}{x})y = e^x[br]$

Sorry the latex fracs are a bit dodgy haha, but I don't know what to do from there.. Have I made any mistakes or can I just not spot the product rule bit?

Teensy typo. :-)

Still more helpful than you.

thank you

Still more helpful than you.

I think the integrating factor is $e^{x + \ln x} = xe^x$. The user has just done some abuse of notation. :-)

Play nice.

nah, mate I realise and edited it because it looked iffy.

That's not my style.