Find the general sollution to [math](x+y)\frac{dy}{dx}=x^2 + xy +1[/math] using the substitution of [math]y=v-x[/math]

Using that [math]\frac{dy}{dx}=\frac{dv}{dx}-1[/math] I plugged that in to get:[math]v(\frac{dv}{dx}-1)=x^2 + x(v-x)+1[/math]. Which rearanges to [math]\frac{dv}{dx}-\frac{1}{v}=x+1[/math]. But from that point I don't know where I'm going.

Could anyone give me a nudge in the right direction?

You'll have to use implicit differentation.

I fear that my typo made your awnser irrelevant, mathworld's article seems to only correlate with how I incorectly typed my work in progress. Sorry for wasting your time.

 

Do you have any ideas as to what I should be doing now?

Well, since your equation of [math](x+y)=x^2+xy+1[/math] have both x, y on the both sides of the equation, you have to use implicit differentation?

 

To use implicit differentation, you do this: [math](x+y)\frac{dy}{dx}=\frac{d}{dx}(x^2+xy+1)[/math].

 

You can find better webpages on implicit differentation than the one you found. If you still don't understand, then I will try to explain it better next time. I don't have the time right now because my girlfriend is here with me.

I may be mistaken but if [math](x+y)\frac{dy}{dx}=x^2 + xy +1[/math] then how can [math](x+y)\frac{dy}{dx}=\frac{d}{dx}(x^2+xy+1)[/math] also?

you can't...

 

first try subbing in v-x for all of the y's and plug in the derivative v-x= y

 

EDIT: woops, you already did that.

 

 

are you familiar with integrating factors and exact equations?

 

or actaully it looks like if you sub in for y again at the end you'll get an answer.