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# C4 Differential Equations Question watch

1. This question is from the January 2011 C4 Paper, I'll skip part a) because it is not required for what I have a problem with.

b) Hence find the integral of 5/[(x-1)(3x+20)]

c) Find the particular solution of the differential equation:
(x-1)(3x+2)[dy/dx] = 5y
For which x=2 at y=8, give your answer in the form y=f(x)

So, for part b) I got the correct answer, being ln(x-1/3x+2) + C

For part c), however, I do not understand why one answer is invalid (although the two results are completely different things).

My working is:
[int.]1/y dy = [int.] 5/[(x-1)(3x+20)] dx
Which gives:
lny = ln(x-1/3x+2) + C

So, cancelling Natural Logs;
y = (x-1)/(3x+2) + D, where D is some new constant.
using the coordinates (2,8), I calculate D to be equal to 63/4.
And so, my final answer is y = (x-1)/(3x+2) + 63/4.

I do not understand why this is incorrect, even though it clearly is.
he method in the markscheme is identical other than it replaces C with a constant Ln(A), and writes instead:
y = A(x-1)/(3x+2)
by using laws of logs.
Could anyone please help me understand why you must replace C with Ln(A) and simply writing it as some new constant gives the incorrect answer?

2. (Original post by JLegion)
T

My working is:
[int.]1/y dy = [int.] 5/[(x-1)(3x+20)] dx
Which gives:
lny = ln(x-1/3x+2) + C

So, cancelling Natural Logs;
y = (x-1)/(3x+2) + D, where D is some new constant.
.
You can't cancel logs like that, which is why you need to move the C to inside the ln, so you can compare the things inside the brackets.
3. (Original post by JLegion)
This question is from the January 2011 C4 Paper, I'll skip part a) because it is not required for what I have a problem with.

b) Hence find the integral of 5/[(x-1)(3x+20)]

c) Find the particular solution of the differential equation:
(x-1)(3x+2)[dy/dx] = 5y
For which x=2 at y=8, give your answer in the form y=f(x)

So, for part b) I got the correct answer, being ln(x-1/3x+2) + C

For part c), however, I do not understand why one answer is invalid (although the two results are completely different things).

My working is:
[int.]1/y dy = [int.] 5/[(x-1)(3x+20)] dx
Which gives:
lny = ln(x-1/3x+2) + C

So, cancelling Natural Logs;
y = (x-1)/(3x+2) + D, where D is some new constant.
using the coordinates (2,8), I calculate D to be equal to 63/4.
And so, my final answer is y = (x-1)/(3x+2) + 63/4.

I do not understand why this is incorrect, even though it clearly is.
he method in the markscheme is identical other than it replaces C with a constant Ln(A), and writes instead:
y = A(x-1)/(3x+2)
by using laws of logs.
Could anyone please help me understand why you must replace C with Ln(A) and simply writing it as some new constant gives the incorrect answer?

You can't cancel logs or antilog unless everything on both sides is inside a log. You can just find out the value of c by subbing in the values of 2 and 8 that they gave you, or you can change c to LnA, rearrange the equation and then sub in, but you can't cancel logs straight away.
4. Ah I see! Thank you for your replies and help.

Ln(y) = Ln(x) + C
Which becomes:
e^ln(y) = e^ln(x) + e^C *
Which becomes:
Y = x + D

* My mistake was at the asterisk.
It should've been:
e^ln(y) = e^[ln(x)+C]
y = e^ln(x) • e^C

And then I could've wrote e^C as D.

Is this correct?

5. Yes, correct.

(Original post by JLegion)
Ah I see! Thank you for your replies and help.

Ln(y) = Ln(x) + C
Which becomes:
e^ln(y) = e^ln(x) + e^C *
Which becomes:
Y = x + D

* My mistake was at the asterisk.
It should've been:
e^ln(y) = e^[ln(x)+C]
y = e^ln(x) • e^C

And then I could've wrote e^C as D.

Is this correct?

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