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\mu \left( \frac{1}{r} \frac{d}{dr} \left( r \frac{dV_0}{dr} \right) - \frac{V_0}{r^2} \right) = 0 \\ \frac{d^2V_0}{dr^2} + \frac{1}{r} \frac{dV_0}{dr} - \frac{V_0}{r^2} = 0 \quad Euler \quad V_0 = Ar^m \\ m = \pm 1 \\ Solution \\ V_0 = Ar + \frac{B}{r} \\ b) B.C. \quad At \quad r = a \quad V_0 = aw, \quad r = b \quad V_0 = bw \\ w_1 = w_2 = w \\ V_0 = aw \\ V_0 = 2aw \\ b = 2a \\ c) \\ V_0 = Ar + \frac{B}{r} \\ aw = Aa + \frac{B}{a} \\ 2aw = 2Aa + \frac{B}{2a} \\ V_0 = wr \\ Subtract \\ 0 = B \left( \frac{1}{a} - \frac{1}{2a} \right) \\ B = 0 \\ A = w \\ aw = Aa + \frac{B}{a}

          \mu \left( \frac{1}{r} \frac{d}{dr} \left( r \frac{dV_0}{dr} \right) - \frac{V_0}{r^2} \right) = 0 \\
\frac{d^2V_0}{dr^2} + \frac{1}{r} \frac{dV_0}{dr} - \frac{V_0}{r^2} = 0 \quad Euler \quad V_0 = Ar^m \\
m = \pm 1 \\
Solution \\
V_0 = Ar + \frac{B}{r} \\
b) B.C. \quad At \quad r = a \quad V_0 = aw, \quad r = b \quad V_0 = bw \\
w_1 = w_2 = w \\
V_0 = aw \\
V_0 = 2aw \\
b = 2a \\
c) \\
V_0 = Ar + \frac{B}{r} \\
aw = Aa + \frac{B}{a} \\
2aw = 2Aa + \frac{B}{2a} \\
V_0 = wr \\
Subtract \\
0 = B \left( \frac{1}{a} - \frac{1}{2a} \right) \\
B = 0 \\
A = w \\
aw = Aa + \frac{B}{a}

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μ( (1)/(r) (d)/(dr) ( r (dV0)/(dr) ) - (V0)/(r^2) ) = 0

(d^2V0)/(dr^2) + (1)/(r) (dV0)/(dr) - (V0)/(r^2) = 0 Euler V0 = Ar^m

m = ±1

Solution

V0 = Ar + (B)/(r)

b) B.C. At r = a V0 = aw, r = b V0 = bw

w1 = w2 = w

V0 = aw

V0 = 2aw

b = 2a

c)

V0 = Ar + (B)/(r)

aw = Aa + (B)/(a)

2aw = 2Aa + (B)/(2a)

V0 = wr

Subtract

0 = B ( (1)/(a) - (1)/(2a) )

B = 0

A = w

aw = Aa + (B)/(a)

Added by Jose Carlos W.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Dominique Jan Tan

01/18/2024

Step 1

C. 4 + D = A Vo = bw Ug = 2av Show more…

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Thanks for your feedback!

I need help understanding this math they're doing. I give up, nothing is ever explained. Pretty sure it's some rule for systems of equations but I'm not a math wizard so I forgot. I just wanna know how they did the math to get the final U_(theta) value. I'm lost after he substitutes b = 2a into the U_theta. I thought they just subtracted the system of equations but then where does the 4*a come from in the denominator of the combined equation? and THEN, what are they even doing under the "subtract" term, seriously I have no semblance of what's going on after we get our system of equations c = [-bp + dVg - Vo = 0 tuLs Ng - AI d3No m = +I d - tio B Vo = A1 + W, Tu2Fw 1 Vo-au. Vo = 0 w B.C. 4 + D = A b) Vo = bw. Ug = 2av 9 = A b = 2a Ug = A1 t c Subtract Qw = Aa + a Qu = Ac + B B = 0 2aW z 2Aa 40 24 4 = w U0 = w r

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