Solve the given differential equation

• chwala
In summary, the constant may be manipulated and it does not matter where one places the constant in these kind of problems.
chwala
Gold Member
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,

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

So, you don't have a problem, right?

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.

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.

chwala
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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