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# P3 differentiation watch

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1. i've been doing questions from the revision guide and have come across 'first order differential equation'..is this just another way of saying differential equation?

anyway, i'm having trouble with these two

1) Given that y^2 = Asin3x, form a first order differential equation that does not contain the constant A

2) Given that ky = 4^x, find a first order differential equation which does not contain the constant k

thank you! x
2. 'First order' means the highest derivative is of order 1 (ie. only dy/dx). If it was a second order equation, there'd be a d^2y/dx^2 in there.

I'm not sure how to get rid of those constants :/.
3. first order differential equation means equation with only dy/dx in it (no d2y/dx2 or so on)...it can obviously contain y, y^2, x x^2 etc (however, we call those non-linear if they have y^2 in them)

looking at number 2:

ky = 4^x

ln ky = ln 4^x
ln k + ln y = xln 4
ln y = xln 4 - ln k
d(ln y)/dx = ln4
dy/dx * 1/y = ln4
dy/dx = yln4

there...does that count?

you can do something similar with 1 maybe:

y^2 = Asin3x
2lny = ln A + ln (sin3x)
dy/dx * 1/y = d( ln(sin3x)/2 )/dx

i dunno how to differentiate the horrid mess on the right (probably by chain rule becomes 3cos3x/2sin3x or something similar )

Edit: I stand by my answer now:

dy/dx = y3cos3x/(2sin3x)
4. I can only get the first equation to:

3acos3x/(2(asin3x)^1/2)
5. (Original post by Bhaal85)
I can only get the first equation to:

3acos3x/(2(asin3x)^1/2)
Yeah, that was the problem I had :/.
6. yeh i cant get rid of the constants!

1) 3y^2cos3x - 2y dy/dx sin3x = 0

2) dy/dx - yln4 = 0
7. yay i got them right!!!!

my answer to question one is equivilent (put the dy/dx bit on the right hand side, and divide through by your ysin3x thingum , and you get the same answer as me!
8. thank you i get no2, but could you explain the first one again pls. thank you so much!
xx
9. ok:

y^2 = Asin3x

ln y^2 = ln (Asin3x)

2lny = ln A + ln (sin3x)

so ln y = lnA/2 + ln(sin3x)/2

differential of ln y with respect to x = dy/dx * 1/y
the LnA/2 disappears (constant)
the ln(sin3x)/2 differentiated...well chain rule:

(1/sin3x * 3cos3x) / 2

so dy/dx * 1/y = 3cos3x/2sin3x

so y * dy/dx = y^2 * 3cos3x/2sin3x

Y * dy/dx * 2sin3x = y^2 * 3cos3x

10. thank you!
11. (Original post by jazzyj)
yeh i cant get rid of the constants!

1) 3y^2cos3x - 2y dy/dx sin3x = 0

2) dy/dx - yln4 = 0
Why isn't the first answer 3ycos3x - 2 dy/dx sin3x = 0?
To include the case when y = 0?

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