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given that y=pi/4 when x=1 solve the differential equation

dy/dx = x sec y cosec^3y

....i've rearranged to get an integral but i'm struggling now

i have cos y sin^3y dy = x dx

help please?

dy/dx = x sec y cosec^3y

....i've rearranged to get an integral but i'm struggling now

i have cos y sin^3y dy = x dx

help please?

Original post by pumpkinpiex

given that y=pi/4 when x=1 solve the differential equation

dy/dx = x sec y cosec^3y

....i've rearranged to get an integral but i'm struggling now

i have cos y sin^3y dy = x dx

help please?

dy/dx = x sec y cosec^3y

....i've rearranged to get an integral but i'm struggling now

i have cos y sin^3y dy = x dx

help please?

Hint: differentiating sin y gives cos y

i'm thinking integration by parts? ..the answer i'm supposed to get is

sin^4y = 2x^2 + k

wheni get that i'll be able to do it

sin^4y = 2x^2 + k

wheni get that i'll be able to do it

Original post by pumpkinpiex

i'm thinking integration by parts? ..the answer i'm supposed to get is

sin^4y = 2x^2 + k

wheni get that i'll be able to do it

sin^4y = 2x^2 + k

wheni get that i'll be able to do it

$\frac{\mathrm{d}y}{\mathrm{d}x} = x \sec y \, \mathrm{cosec}^3 y[br][br]\Rightarrow \; \cos y \sin^3 y \; \mathrm{d}y = x \; \mathrm{d} x [br][br]\Rightarrow \; \int \cos y \sin^3 y \; \mathrm{d}y = \int x \; \mathrm{d} x$

The RHS should be very simple to integrate; but look at the LHS. And think of what differentiates to give it - this sort of simple 'recognition' of what differentiates to give what saves you a nasty Integration by Parts job:

$\Rightarrow \; \sin^4 y = \frac{x^2}{2} + c$

And then you sub in your numbers to get $c = \frac{-1}{4}$.

Original post by AmroTT

$\frac{\mathrm{d}y}{\mathrm{d}x} = x \sec y \, \mathrm{cosec}^3 y[br][br]\Rightarrow \; \cos y \sin^3 y \; \mathrm{d}y = x \; \mathrm{d} x [br][br]\Rightarrow \; \int \cos y \sin^3 y \; \mathrm{d}y = \int x \; \mathrm{d} x$

The RHS should be very simple to integrate; but look at the LHS. And think of what differentiates to give it - this sort of simple 'recognition' of what differentiates to give what saves you a nasty Integration by Parts job:

$\Rightarrow \; \sin^4\ y = \frac{x^2}{2} + c$

And then you sub in your numbers to get $c = \frac{-1}{4}$.

The RHS should be very simple to integrate; but look at the LHS. And think of what differentiates to give it - this sort of simple 'recognition' of what differentiates to give what saves you a nasty Integration by Parts job:

$\Rightarrow \; \sin^4\ y = \frac{x^2}{2} + c$

And then you sub in your numbers to get $c = \frac{-1}{4}$.

Would it not be $\frac{1}{4} sin^4 y$ on the LHS?

How did you manage to do this in the end?

Original post by rggregrgr

Would it not be $\frac{1}{4} sin^4 y$ on the LHS?

Yeah it would be, he's was just saying try and differentiate $sin^4y = \frac{1}{2}x^2 + C$ and see what adjustments have to be made e.g. coefficients.

Original post by soutioirsim

Yeah it would be, he's was just saying try and differentiate $sin^4y = \frac{1}{2}x^2 + C$ and see what adjustments have to be made e.g. coefficients.

Ah apologies! Thanks.

Original post by rggregrgr

Would it not be $\frac{1}{4} sin^4 y$ on the LHS?

Mother*$&#ing indices.

Indeed it would, apologies.

Original post by rggregrgr

Ah apologies! Thanks.

Original post by AmroTT

Mother*$&#ing indices.

Indeed it would, apologies.

Indeed it would, apologies.

Nevermind

Original post by AmroTT

$\frac{\mathrm{d}y}{\mathrm{d}x} = x \sec y \, \mathrm{cosec}^3 y[br][br]\Rightarrow \; \cos y \sin^3 y \; \mathrm{d}y = x \; \mathrm{d} x [br][br]\Rightarrow \; \int \cos y \sin^3 y \; \mathrm{d}y = \int x \; \mathrm{d} x$

The RHS should be very simple to integrate; but look at the LHS. And think of what differentiates to give it - this sort of simple 'recognition' of what differentiates to give what saves you a nasty Integration by Parts job:

$\Rightarrow \; \sin^4 y = \frac{x^2}{2} + c$

And then you sub in your numbers to get $c = \frac{-1}{4}$.

The RHS should be very simple to integrate; but look at the LHS. And think of what differentiates to give it - this sort of simple 'recognition' of what differentiates to give what saves you a nasty Integration by Parts job:

$\Rightarrow \; \sin^4 y = \frac{x^2}{2} + c$

And then you sub in your numbers to get $c = \frac{-1}{4}$.

can you explain how this helps please ? Just im quite confused following through :/

Original post by todsy

Original post by AmroTT

$\frac{\mathrm{d}y}{\mathrm{d}x} = x \sec y \, \mathrm{cosec}^3 y[br][br]\Rightarrow \; \cos y \sin^3 y \; \mathrm{d}y = x \; \mathrm{d} x [br][br]\Rightarrow \; \int \cos y \sin^3 y \; \mathrm{d}y = \int x \; \mathrm{d} x$

The RHS should be very simple to integrate; but look at the LHS. And think of what differentiates to give it - this sort of simple 'recognition' of what differentiates to give what saves you a nasty Integration by Parts job:

$\Rightarrow \; \frac{\sin^4 y}{4} = \frac{x^2}{2} + c$

And then you sub in your numbers to get $c = \frac{-7}{16}$.

The RHS should be very simple to integrate; but look at the LHS. And think of what differentiates to give it - this sort of simple 'recognition' of what differentiates to give what saves you a nasty Integration by Parts job:

$\Rightarrow \; \frac{\sin^4 y}{4} = \frac{x^2}{2} + c$

And then you sub in your numbers to get $c = \frac{-7}{16}$.

can you explain how this helps please ? Just im quite confused following through :/

So those are the corrected numbers, integrated properly. That now gives you a general solution, which is your answer, unless they've given you other values to substitute in...?

Original post by AmroTT

So those are the corrected numbers, integrated properly. That now gives you a general solution, which is your answer, unless they've given you other values to substitute in...?

thank you !! lol cos i didn't understand this I kind of trawled around the book and now got it under my belt

Original post by pumpkinpiex

given that y=pi/4 when x=1 solve the differential equation

dy/dx = x sec y cosec^3y

....i've rearranged to get an integral but i'm struggling now

i have cos y sin^3y dy = x dx

help p

dy/dx = x sec y cosec^3y

....i've rearranged to get an integral but i'm struggling now

i have cos y sin^3y dy = x dx

help p

Original post by rggregrgr

Would it not be $\frac{1}{4} sin^4 y$ on the LHS?

i thought when you integrate cos y sin^3 y you get 1/4 sin y x

Original post by todsy

thank you !! lol cos i didn't understand this I kind of trawled around the book and now got it under my belt

Does anyone know how to answer this question i don't see how to integrate get 1/4 sin^4 y

Original post by ADOOLEY

i thought when you integrate cos y sin^3 y you get 1/4 sin y x

Does anyone know how to answer this question i don't see how to integrate get 1/4 sin^4 y

Does anyone know how to answer this question i don't see how to integrate get 1/4 sin^4 y

Like i get how sin^3 y integrates to give 1/4 sin^4 y but what happens to the cos y

Original post by ADOOLEY

Like i get how sin^3 y integrates to give 1/4 sin^4 y but what happens to the cos y

Its 10 years old. Its better to start your own thread rather than resurrect old, zombie ones. But you can use substitution u = sin(y) or simply reverse chain rule to get

Int cos(y) sin^3(y) dy -> 1/4 sin^4(y) + c

Original post by mqb2766

Its 10 years old. Its better to start your own thread rather than resurrect old, zombie ones. But you can use substitution u = sin(y) or simply reverse chain rule to get

Int cos(y) sin^3(y) dy -> 1/4 sin^4(y) + c

Int cos(y) sin^3(y) dy -> 1/4 sin^4(y) + c

I thought when you integrate cos y you get -siny

Original post by ADOOLEY

I thought when you integrate cos y you get -siny

https://www.wolframalpha.com/input?i=integrate+cos%28y%29

When you differentiate sin(y) what do you get? If youre confused just sketch the curves and make sure you get the signs right.

- Edexcel A2 Mathematics: Core Mathematics C4 6666 01 - 22 June 2018 [Exam Discussion]
- A level maths UMS marks!!!
- Best resources for OCR MEI maths Alevel or maths in general
- Any revision tips/materials for Bio and maths?
- Worst exam boards ever...help??
- How do I self study for the OCR FSMQ?
- MEI numerical methods help
- MEI numerical methods revision help
- AQA Art GCSE themes 2024
- Econ+Maths joint honours, or straight Econ?
- differentiation/integration
- A-Level Mathematics Study Group 2024-25
- A-level History Study Group 2022-2023
- Chemistry A-level
- what is core maths?
- ocr level 3 additional maths
- ocr trapezium rule question
- Advice for Year 12
- history ocr HELP PLEASE
- Resources that teach you A level courses from scratch?

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