Solve the given differential equation

In summary, the constant may be manipulated and it does not matter where one places the constant in these kind of problems.
  • #1
chwala
Gold Member
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Homework Statement
See attached.
Relevant Equations
separation of variables
I am on differential equations today...refreshing.

Ok, this is a pretty easier area to me...just wanted to clarify that the constant may be manipulated i.e dependant on approach. Consider,

1691324528542.png


1691324637563.png


Ok I have,

##\dfrac{dy}{6y^2}= x dx##

on integration,

##-\dfrac{1}{6y} + k = \dfrac{x^2}{2}##

##k= \dfrac{x^2}{2} + \dfrac{1}{6y}##

using ##y(1)=0.04## we shall get,

##k=\dfrac{28}{6}##

##\dfrac{28}{6}-\dfrac{x^2}{2}=\dfrac{1}{6y}##

...

aaargh looks like i will get the same results...cheers
 
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  • #3
Mark44 said:
So, you don't have a problem, right?
Correct, I do not have a problem.

Where one places the constant doesn't really matter in these kind of problems.
 
  • #4
chwala said:
Where one places the constant doesn't really matter in these kind of problems.
It does matter in certain cases.
If I worked the problem like this ...
##\frac {dy}{y^2} = 6x~dx##
##\frac{-1}y = 3x^2## (Omitting the constant for now)
##y = \frac{-1}{3x^2} + C## (Adding the constant now)

The above gives the wrong value for C when you substitute in the initial condition.
I'm not saying that's what you meant. In the second line above, you could write it as ##\frac{-1}y = 3x^2 + C_1## or as ##\frac{-1}y + C_2= 3x^2##.
The constants will be different, but when you solve for y in either equation, the solutions will be the same.
chwala said:
##\dfrac{28}{6}-\dfrac{x^2}{2}=\dfrac{1}{6y}##
You should write the solution in the form y = f(x), as was shown in the solution.
 
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  • #5
Mark44 said:
It does matter in certain cases.
If I worked the problem like this ...
##\frac {dy}{y^2} = 6x~dx##
##\frac{-1}y = 3x^2## (Omitting the constant for now)
##y = \frac{-1}{3x^2} + C## (Adding the constant now)

The above gives the wrong value for C when you substitute in the initial condition.
I'm not saying that's what you meant. In the second line above, you could write it as ##\frac{-1}y = 3x^2 + C_1## or as ##\frac{-1}y + C_2= 3x^2##.
The constants will be different, but when you solve for y in either equation, the solutions will be the same.

You should write the solution in the form y = f(x), as was shown in the solution.
@Mark44 I realized that it would just give the same solution, thus left it hanging.
 

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