Solving Exact Diff. Eqns w/ Coefficients >1 in Soln: dr/dΘ=(r^2 sin Θ)/(2r cos Θ -1)

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gw1500se

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Solving Exact Diff. Eqns w/ Coefficients >1 in Soln: dr/dΘ=(r^2 sin Θ)/(2r cos Θ -1)

I am trying to refresh my diffy and am having trouble recalling the handling of coefficents >1 in the solution. For example, given the exact differential equation:

dr/dΘ=(r^2 sin Θ)/(2r cos Θ -1)

Since it is exact I get the form Mdr + NdΘ = 0:

(2r cos Θ -1)dr-(r^2 sin Θ)dΘ = 0

Solving, I get:

2r^2 cos Θ -r = c

My book says the answer is:

r^2 cos Θ -r = c

I understand the concept of incorporating coefficients on the left into the constant on the right. However, in this case I don't see how the coefficient '2' can be incorporated from just the one term. How can a constant 'c' be obtained that does not change the relationship between the two terms? Can someone please explain? TIA.
 
Last edited:
gw1500se said:
I am trying to refresh my diffy and am having trouble recalling the handling of coefficents >1 in the solution. For example, given the exact differential equation:

dy/dΘ=(r^2 sin Θ)/(2r cos Θ -1)

Since it is exact I get the form Mdy + NdΘ = 0:

(2r cos Θ -1)dy-(r^2 sin Θ)dΘ = 0

Solving, I get:

2r^2 cos Θ -r = c

My book says the answer is:

r^2 cos Θ -r = c

I understand the concept of incorporating coefficients on the left into the constant on the right. However, in this case I don't see how the coefficient '2' can be incorporated from just the one term. How can a constant 'c' be obtained that does not change the relationship between the two terms? Can someone please explain? TIA.
Click to expand...
Please show the steps taken for the solution.
 
gw1500se said:
I am trying to refresh my diffy and am having trouble recalling the handling of coefficents >1 in the solution. For example, given the exact differential equation:

dy/dΘ=(r^2 sin Θ)/(2r cos Θ -1)

Since it is exact I get the form Mdy + NdΘ = 0:

(2r cos Θ -1)dy-(r^2 sin Θ)dΘ = 0

Solving, I get:

2r^2 cos Θ -r = c

My book says the answer is:

r^2 cos Θ -r = c

I understand the concept of incorporating coefficients on the left into the constant on the right. However, in this case I don't see how the coefficient '2' can be incorporated from just the one term. How can a constant 'c' be obtained that does not change the relationship between the two terms? Can someone please explain? TIA.
Click to expand...
Is the given equation:

dr/dΘ=(r^2 sin Θ)/(2r cos Θ -1)
 
Subhotosh Khan said:
Is the given equation:

dr/dΘ=(r^2 sin Θ)/(2r cos Θ -1)
Click to expand...

Yes, that was a typo. Sorry.

Here are my steps finding the anti-derivatives of M and N:

(r^2 cos Θ -r) - (r^2 (-cos Θ)) = c
r^2 cos Θ -r + r^2 cos Θ = c
2r^2 cos Θ -r = c
 
gw1500se said:
I am trying to refresh my diffy and am having trouble recalling the handling of coefficents >1 in the solution. For example, given the exact differential equation:

dr/dΘ=(r^2 sin Θ)/(2r cos Θ -1)

Since it is exact I get the form Mdr + NdΘ = 0:

(2r cos Θ -1)dr-(r^2 sin Θ)dΘ = 0

Solving, I get:

2r^2 cos Θ -r = c

My book says the answer is:

r^2 cos Θ -r = c

I understand the concept of incorporating coefficients on the left into the constant on the right. However, in this case I don't see how the coefficient '2' can be incorporated from just the one term. How can a constant 'c' be obtained that does not change the relationship between the two terms? Can someone please explain? TIA.
Click to expand...

The problem is not the incorporation of constants but the integration and the particular assumption of separability. Ignoring this problem for a moment, suppose r = 1/cos(\(\displaystyle \theta\)). What is the integral of 2r cos Θ? On the other hand, what if r=sin(\(\displaystyle \theta\)). What is the integral of 2r cos Θ?

To put this in more general terms
\(\displaystyle \int \, u'\, v\, =\, u v \,-\, \int\, u\, v'\)
If you want to approach this problem as separable the way it stands, you are forgetting the second part of the equation on the right hand side when you integrate.
 
Unless I am totally misunderstanding I don't think we are talking about the same thing with respect to solving exact differential equations. The theorem being used here involves partial derivatives.

In part (assuming an exact differential equation):

Given the equation M dx + N dy = 0 = dU, then U exists such that δU/δx=M and δU/δy=N

Thus we are finding the anti-derivative of M and N, not integrating them.
 
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