# FP1 first order differentials question June 04 q.6bWatch

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#1
Argh, back again :/

In part a I found the general solution of this curve:

(A) dy/dx + 2y = x

to be

(B) y = 0.5x - 0.25 + ce^(-2x)

Where C is a constant (I kept it in fractions but it's easier to type decimals on here) which is right, according to my answers.

Then part b tells me that y = 1 when x = 0, and asks me to find the exact coordinates of the minimum point of the curve.

Now, am I going about this in the right way in that I sub in y=1 and x=0 into equation (B) to find C (which I find to be a quarter) and then differentiate equation (B) (now knowing C) and solve equal to zero? Because when I try doing that I end up getting 0 for both y and x, when the answers are

x = 0.5(ln5)
y = 0.25(ln5)

:/

Just wondering if I was going wrong in my method because I can't find any mistakes I've made with the number crunching.. Thanks!

ps. is there any way of doing the fancy equation thing on here easier than learning the code??
0
11 years ago
#2
dy/dx = x - 2y

Find the constant, C.
Next, you know what y is. sub that in to the equation above and set to 0.

Edit: just looking at it, your method should work. let me see...
0
11 years ago
#3
Your C value doesn't seem right.
You get:

1= -0.25+c => c= 1.25
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#4
Oh lol, I'm such an idiot. I got that but then got the final value wrong *smacks forehead*

Okay I've done that and got the x value right, but the y value is coming out wrong :/ can you show me how you got the coordinates?
0
11 years ago
#5
y(0)=1 = 0 - 1/4 + c => c = 5/4

also,

y = 0.25(ln5)
x = 0.5(ln5)
0
11 years ago
#6
get y' then substitute for c = 5/4 and solve for x
0
11 years ago
#7
0.5 - 5/2.e^(-2x) = 0

1 = 5e^(-2x)
-ln5 = -2x
x= 0.5ln5

from the other equation

dy/dx + 2y = x
0 + 2y = 0.5ln5
y = 0.25ln5

Have you got your y and x the right way round? My answers dont match.
0
#8
Yeah I have the x coordinate, the y one keeps coming out wrong:

x = 0.5(ln5)

y = 0.5x - 0.25 + (5/4)e^(-2x)
y = (1/4)(ln5) - (1/4) + (5/4)e(-(ln5))
4y = ln5 - 1 + 5e(ln(1/5))
4y = ln5 - 1 + 1
y...

Wait, I've done it now :P lol, thanks for your help

edit: yeah, ninja, I had them the wrong way around, sorry (edited)
0
11 years ago
#9
(Original post by tekp)
Yeah I have the x coordinate, the y one keeps coming out wrong:

x = 0.5(ln5)

y = 0.5x - 0.25 + (5/4)e^(-2x)
y = (1/4)(ln5) - (1/4) + (5/4)e(-(ln5))
4y = ln5 - 1 + 5e(ln(1/5))
4y = ln5 - 1 + 1
y...

Wait, I've done it now :P lol, thanks for your help

edit: yeah, ninja, I had them the wrong way around, sorry (edited)
You could save tonnes of working by using the dy/dx equation they gave you and subbing in dy/dx=0 and x=0.5ln5.
0
#10
Oh yeah.. I'm so slow, lol. Thanks
0
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