Differential equation help

10i)

http://www.ocr.org.uk/Images/59552-question-paper-unit-4724-01-core-mathematics-4.pdf

For k I got .07438118316

The answer for k is .07438118377

I can't figure out where I have gone wrong...

I got K = Ln [ 3^(-1/3) * 5^(1/3) ] - 0.2310490602

Reply 1

Indeterminate

3

Original post by IShouldBeRevising_

10i)

http://www.ocr.org.uk/Images/59552-question-paper-unit-4724-01-core-mathematics-4.pdf

For k I got .07438118316

The answer for k is .07438118377

I can't figure out where I have gone wrong...

I got K = Ln [ 3^(-1/3) * 5^(1/3) ] - 0.2310490602

Assuming you used the partial fractions form in (i) to do (ii), you may have made an error integrating or using the initial conditions.

Post working :smile:

Reply 2

davros

16

Original post by IShouldBeRevising_

10i)

http://www.ocr.org.uk/Images/59552-question-paper-unit-4724-01-core-mathematics-4.pdf

For k I got .07438118316

The answer for k is .07438118377

I can't figure out where I have gone wrong...

I got K = Ln [ 3^(-1/3) * 5^(1/3) ] - 0.2310490602

Where have you got all those horrible decimals from?

The answer is given in an exact form, so keep everything expressed as ln(something) until you get to the very end and then everything drops out nicely!

Reply 3

IShouldBeRevising_

OP

2

Original post by davros

Where have you got all those horrible decimals from?

The answer is given in an exact form, so keep everything expressed as ln(something) until you get to the very end and then everything drops out nicely!

Yeah I did again but this time kept everything in exact form and got the correct answer, thanks.

Differential equation help

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