Recent content by ilikesoldat

You sound like a hardworking guy to my standards, and you're obviously willing to study for this final. The people on these forums just were exceptionally good students and then feel everyone should be like that. IMO you don't have an attitude or anything like that, just throwing this out there...

I have to think about this OK, your answer does not depend on y at all. Deriving the original equation x^3+y^2=8y, you get y'=(3x^2)/(8-2y-x^3). Try (1,2). On yours it is still y'=-3. On this one it will be y' =2. Logarithmic differentiation, too, provides y'=-3x^2, which is like your dividing...

Hey, I found a thread about part of what I'm trying to ask long ago: https://www.physicsforums.com/threads/implicit-differentiation.178328/ Basically, I noticed that if you multiply by x or by y in an equation before implicitly deriving, you get two different answers. Unfortunately their whole...

That makes sense but it still bugs me to just assume it is a constant :( And to the other person, the question was dy/dx=x*e^(ax^2) solve the differential equation, so yes it would be integral of x*e^ax^2 dx (except you forgot the squared on top of the x in e's exponent) thanks for...

I solved it with a as a constant and got the right answer, but . . . How do i know a is a constant? I can't just assume that can I? What is the proof for a being a constant

Homework Statement dy/dx=x*e^(ax^2) solve the differential equation Homework Equations integral of e^x=e^x The Attempt at a Solution im not really sure how to do it when there are two variables in the exponent? i tried several things like u=x^2 1/2du=xdx then 1/2*int[e^a*u]...

Oh, I thought equivalent meant equal. my bad... This helped a lot thanks a ton. And yeah that makes sense that log of some things doesn't have to be a constant -.-.

But by your method then they WOULD differ and wouldn't be equal, if they differed by a constant...? Also, technically didn't I show that C1=-C2 and C2=-C1? Because ln (1) = C1+C2, so then 0=C1+C2, and then C1=-C2 or -C1=C2 Also, taking the derivative of each would give 0 and 0, since ln of...

Homework Statement Show that the two formulas are equivalent integral [sec x dx] = ln|sec x + tan x| + C integral [sec x dx] = -ln|sec x - tan x| + C Homework Equations Pythagorean ID's? Log rule of addition The Attempt at a Solution Well, I realized the formulas can only be equivalent if...