Missing a 'y' in solution.

[ContraHacker]

Good evening! I am very thankful for this forum. I'm happy how well it works. Please help me with this :



My solution :



I know I've done something wrong. Kindly point it out.

Thanks!

 

[Deleted member 4993]

Good evening! I am very thankful for this forum. I'm happy how well it works. Please help me with this :

View attachment 12915

My solution :

View attachment 12916View attachment 12917

I know I've done something wrong. Kindly point it out.

Thanks!

You wrote:

(2/y) * (dy/dx) = (y/x) + ln(x) * (dy/dx)..........................................correct

Next line should be:

(dy/dx) * [(2/y) - ln(x)] = y/x

I think you are trying to skip steps .... and messing up.

Otherwise excellent work (neat and logical)!!

 

[ContraHacker]

You wrote:

(2/y) * (dy/dx) = (y/x) + ln(x) * (dy/dx)..........................................correct

Next line should be:

(dy/dx) * [(2/y) - ln(x)] = y/x

I think you are trying to skip steps .... and messing up.

Otherwise excellent work (neat and logical)!!

Thanks man! I was only looking for a solution to my problem, but your compliment made my day! I'll be more careful in my exam tommorow.

 

[lookagain]

ContraHacker, one of your main errors is that you writing it wrong as you are using horizontal style
combined with the Order of Operations. You must use adequate grouping symbols. Don't use "X"
or "x" for multiplication. Leave that to grouping symbols and the asterisk.
(Technically speaking, the parentheses around y/x and 2/y are for spacing purposes and
clarification/emphasis, for example.)

Altering the look of Subhotosh Khan's next step to reflect it more with the problem and with function notation:

(dy/dx) * [(2/y) - log(x)] = y/x

dy/dx = (y/x)/[(2/y) - log(x)]

At some point, multiply the numerator and the denominator of the main fraction on the right side of the equals
sign by y:

dy/dx = y*(y/x)/{y*[(2/y) - log(x)]}

dy/dx = (\(\displaystyle \ y^2\)/x)/[2 - ylog(x)]....... edited typo

At some point, multiply each side by x:

x(dy/dx) = [x\(\displaystyle y^2\))/x]/[2 - ylog(x)]

x(dy/dx) =\(\displaystyle \ y^2\)/[2 - ylog(x)]

 

[ContraHacker]

ContraHacker, one of your main errors is that you writing it wrong as you are using horizontal style
combined with the Order of Operations. You must use adequate grouping symbols. Don't use "X"
or "x" for multiplication. Leave that to grouping symbols and the asterisk.
(Technically speaking, the parentheses around y/x and 2/y are for spacing purposes and
clarification/emphasis, for example.)

Altering the look of Subhotosh Khan's next step to reflect it more with the problem and with function notation:

(dy/dx) * [(2/y) - log(x)] = y/x

dy/dx = (y/x)/[(2/y) - log(x)]

At some point, multiply the numerator and the denominator of the main fraction on the right side of the equals
sign by y:

dy/dx = y*(y/x)/{y*[(2/y) - log(x)]}

dy/dx = (\(\displaystyle \ y^2\)/x)[2 - ylog(x)]

At some point, multiply each side by x:

x(dy/dx) = [x\(\displaystyle y^2\))/x]/[2 - ylog(x)]

x(dy/dx) =\(\displaystyle \ y^2\)/[2 - ylog(x)]

Noted. Thanks for your reply. I will be careful about my parenthesis and asterisks from now on. Multiplying y to the main fraction on RHS solved the question.