Particular Integration

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haiderali

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Nov 13, 2013
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6
Im trying to solve this
y' = sqrt(y/x)

whith [y(1) = 4]


i have found the integration ... and its [ y = x + (c2/4)] or [y = (sqrt(x) + (c/2))]

but the problem is with the C value ... when i try to find it .... it gives me a different vlue than the value which is on the book answers ..

the book answer is [y = (sqrt(x) + 1)2]


i have already done many others particular Integrals .. in which the C was not touched .. i mean that C was just a simple C ... not a C2 ... and there the answers were right ...but not here ..


Can some one help me ;)
im a newbie ... sorry for my english ...


edit : Can any one tell me how to find the value of ""t"" in this equation :

ln(sqrt(1/2t)) = ( ln|x| ) + c ----> the same problem ... i dont understand if when i raise the ln equation to "e" .. do i have to raise the C too .. or not ..

do i have to do like :
eln(sqrt(1/2t)) = eln|x| + ec

or

eln(sqrt(1/2t)) = eln|x| + c ?????
how do i have to procede to get the "t" value :(

im getting mad .. help
 
Last edited:
I don't know about others, but I've never heard of "radq". What does this mean?

Thank you! ;)
 
haiderali said:
Im trying to solve... y' = sqrt(y/x), with y(1) = 4.

i have found... y = x + (c2/4) or y = (sqrt(x) + (c/2))
Click to expand...
These two functions are not the same! Also, neither appears to be correct. Please reply showing your steps for at least one of your answers. Thank you! ;)
 
haiderali said:
Im trying to solve this
y' = sqrt(y/x)
Click to expand...

\(\displaystyle \frac{dy}{dx}= \sqrt{\frac{y}{x}}= \frac{y^{1/2}}{x^{1/2}}= y^{1/2}x^{-1/2}\)
\(\displaystyle y^{-1/2}dy= x^{-1/2}dx\)

whith [y(1) = 4]


i have found the integration ... and its [ y = x + (c2/4)] or [y = (sqrt(x) + (c/2))]
Click to expand...
No, thats wrong. You don't show how you got that so we cannot say what you might have done wrong. I suspect it is something as simple as mistakenly thinking that \(\displaystyle (a+ b)^2= a^2+ b^2.

but the problem is with the C value ... when i try to find it .... it gives me a different vlue than the value which is on the book answers ..
Click to expand...
Again, you don't show what you did so I don't know what you did wrong.

the book answer is [y = (sqrt(x) + 1)2]


i have already done many others particular Integrals .. in which the C was not touched .. i mean that C was just a simple C ... not a C2 ... and there the answers were right ...but not here ..


Can some one help me ;)
im a newbie ... sorry for my english ...


edit : Can any one tell me how to find the value of ""t"" in this equation :

ln(sqrt(1/2t)) = ( ln|x| ) + c ----> the same problem ... i dont understand if when i raise the ln equation to "e" .. do i have to raise the C too .. or not ..

do i have to do like :
eln(sqrt(1/2t)) = eln|x| + ec

or

eln(sqrt(1/2t)) = eln|x| + c ?????
how do i have to procede to get the "t" value :(

im getting mad .. help[/QUOTE]\)
 
Last edited:
\(\displaystyle y(1) = 4\) ???

I could understand something like \(\displaystyle y(1) =\) the value \(\displaystyle 1\) plugged into the x values of the equation. Or \(\displaystyle f(1) =\) the value \(\displaystyle 1\) plugged into the x values of the equation.
 
Last edited:
its like 4 = quation in which you have put 1 as X values to find the C value as the result.

exmp.

equation --> y = 3x2 + 2x+c
condition to find c Value --> y(1) = 4

finding the value of c :

4 = 3*(12) +2*1 +c

c= 4-3-2
c=-1
 
Ok, I see. He was giving the answer, and then asking for the question.
 
i want to use the same procedure to do in this equation ..
but i dont know how to get the c value ..

y' = sqrt(y/x)

whith [y(1) = 4]
 
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