Question

Can someone confirm my calculus answer regarding integrals? y'+(2x-1)y=0

I got y=e^(-x^2+x)

Answer 1

gotcha. I'm running into a problem though, it's supposed to satisfy a "boundary condition" (not entirely sure what that is) of y(1)=2, but when I plug in x=1 I get y=1

Answer 2

I think we missed C but C is in the front of e
I mean \(y = Ce^{-x^2+x}\) so, if y (1) =2, just plug x =1, y =2 into the solution
\[2=Ce^{-1^2+1}=Ce^0= C\] therefore C =2
Now, plug it back to the solution
\[y = Ce^{-x^2+x}= 2e^{-x^2+x}\]

Answer 3

why is it C*e instead of C+e

Answer 4

also if we fill in x with 1, wouldn't we get Ce^2?

Answer 5

also can you explain how to set this up? http://puu.sh/9gqir/f2a80deb8e.png

Answer 6

ok,
y'= -(2x-1)y -->\(\dfrac{dy}{dx}=-(2x-1)y\rightarrow \dfrac{dy}{y}=(-2x +1)dx\)
Integral both sides
\[lny = -x^2+x +C\]
take e both sides
\[\Large y = e^{the~ whole~ thing}= e^{-x^2+x+C}= e^C*e^{-x^2+x}\]
e^C = C because it is a constant so that C * e^(-x^2+x} not + as usual

Answer 7

for the previous one, re read my comment
\[e^{-x^2+x}= e^0\]
the minus sign is not include in x^2 --> -x^2 = -1^2 , then +1 =0

Answer 8

for the next problem , you need to write out the equation of the distance first, then take derivative of that function to get the velocity function. That's the way you make up ODE

Answer 9

A little late.
Refer to a solution using Mathematica 9.