Really Stuck With Some Calculus...

demon8991

demon8991

demon8991

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Joined
7 Jan 2003
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Gold Coast, Australia
Hello been working through 30 questions in prepartion for an exam. But theres 3 i cant do and they are driving me mad.

Would any of you be able to help?

Use an Integrating Factor to solve the following equations for y:

dy/dx + y/tanx=2 with the condition y(pi/4)=sqrt2 - 3



Solve the following homogeneous equations for y:

dy/dx = y/x + 2(x/y) where y(1) = sqrt2



dy/dx = y/x + (y/x)^4 where y(1) = 1


Thanks in advance guys.
 
demon8991 said:
Use an Integrating Factor to solve the following equations for y:

dy/dx + y/tanx=2 with the condition y(pi/4)=sqrt2 - 3
Click to expand...

The integrating factor in this question is e^(integral of [1/tanx]), multiply the equation by this factor and you will and you have a perfect derivative on the left hand side. Integrate both sides, put in your boundary conditions and you should have an answer.

demon8991 said:
Solve the following homogeneous equations for y:

dy/dx = y/x + 2(x/y) where y(1) = sqrt2



dy/dx = y/x + (y/x)^4 where y(1) = 1


Thanks in advance guys.
Click to expand...

For both of these questions, make the substitution y=kx and simplify, also write dy/dx in terms of dk/dx and integrate both sides.

Alex
 
Right ok thats what i got earlier but couldnt do the next step,jus to check this is the right equation yer?

e^(integral (1/tanx))dy/dx + (y/tanx)e^(integral (1/tanx)) = 2e^(integral (1/tanx))


Coz i cant see what to do from there.
 
Last edited:
demon8991 said:
Right ok, for the first one does the y value jus dissapear, like is the first line of working just.

e^(integral (1/tanx))dy/dx + (1/tanx)e^(integral (1/tanx)) = 2e^(integral (1/tanx))

or is it

e^(integral (1/tanx))dy/dx + (y/tanx)e^(integral (1/tanx)) = 2e^(integral (1/tanx))
Click to expand...

Nothing dissapears, just multiply the whole equation by the intergrating factor, so the second is correct.

Alex

Edit:

The left side is a perfect derivative d/dx(Iy) where I is the integrating factor.
 
Yer this is the same trap as i got in earlier i get y = 2!

But the question says 'with condition y(pi/4) = sqrt 2 - 3
 
The answer i got is y sin(x) = -2 cos(x) + C, i dont really understand those boundary conditions, it should give a value for y and x at a certain point to find C.
 
Having looked at the first question - it's completely routine and there's no reason to give up! Don't lose hope, just follow the method you've been taught.

For the very first bit:

dy/dx + (1/tanx) y = 2

Integrating factor is exp(Integral((1/tanx) dx)) = sin x

(an easy integral)

so

sin x (dy/dx) + y cos(x) = 2

or

d/dy(y sin x) = 2

so y sin x = 2 x + c

and use your conditions to find c!
 
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