Expert: Paul Klarreich - 12/2/2008

Question
Hi, I hope you could help me with the following questions: (i've provided the answer as the answers given by another expert isn't correct)

1. Differentiate with respect to x:

i) x ln x - x

answer: ln x

2. Integrate with respect to x:

i) (x + 1)/(2x + 1)

answer: 1/2x + 1/4 ln(2x + 1) + C

3. Solve the following differential equations:

i) dy/dx = sin x/cos y
ii) dy/dx = 4y / x(y-3)
iii) dy/dx = x^2 (3-2y)  y(0)=0
iv) dy/dx = (1-y)/(1+x)  y(0)=0

answer:
i)siny + cosx = C  or y = sin^-1 (cos x + C)

ii) x^4y^3 = C e^y

iii) 1/3x^3 + 1/2 ln (3 - 2y) = 0  or
y = 1/2 (3 - e^(2x^3/3))

iv) y = x/(1+x)

Thanks very much!

Answer
Questioner:   Lill
Category:  Advanced Math
Private:  No
 
Subject:  Natural Logarithmic Function (Differentation and Integration)
Question:  Hi, I hope you could help me with the following questions: (i've provided the answer as the answers given by another expert isn't correct)

1. Differentiate with respect to x:

i) x ln x - x

answer: ln x

>> looks good.

2. Integrate with respect to x:

i) (x + 1)/(2x + 1)


2x+1 ) x+1  (1/2
      x+1/2
     ------
       1/2
        1
1/2 + ----------
     2(2x + 1)


answer: 1/2x + 1/4 ln(2x + 1) + C

>> (1/2)x, I assume you mean.  


3. Solve the following differential equations:

i) dy/dx = sin x/cos y

i)siny + cosx = C  or y = sin^-1 (cos x + C)

>> OK.


ii) dy/dx = 4y / x(y-3)


ii) x^4y^3 = C e^y

>> OK.

iii) dy/dx = x^2 (3-2y)  y(0)=0

- ln(3 - 2y)/2 = x^3/3  + C

- ln(3)/2 = + C

C = - ln(3)/2

- ln(3 - 2y)/2 = x^3/3  - ln(3)/2

          3
(1/2)ln(------ ) = x^3/3
       3 - 2y

MAYBE NOT.


iii) 1/3x^3 + 1/2 ln (3 - 2y) = 0  or
y = 1/2 (3 - e^(2x^3/3))



iv) dy/dx = (1-y)/(1+x)  y(0)=0

answer:
dy     dx
---- = ----
1-y    1+x

-ln(1-y) = ln(1+ x) + C

C = 0

-ln(1-y) = ln(1+x)

0 = ln(1+x) + ln(1-y)

(1 + x)(1 - y) = 1,

etc.


iii) 1/3x^3 + 1/2 ln (3 - 2y) = 0  or
y = 1/2 (3 - e^(2x^3/3))

iv) y = x/(1+x)

Thanks very much!

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