Solve the first order differential equation

• chwala
In summary, Mark provided a solution to the equation xy=2x+k, where x=-x^3-4x^2+x and y=-x^1+2x^0.5. After solving for x, he verified that x=-0.5 and therefore the solution is correct.

chwala

Gold Member
Homework Statement
Solve ##x\dfrac {dy}{dx}+y##=##x^{-1/2}##, ##x>0##given that ##y=2## when ##x=4##
Relevant Equations
First order ode
From my working...I am getting,
##xy=####\int x^{-1/2}\ dx##
##y##=##\dfrac {2}{x}##+##\dfrac {k}{x}##
##y##=##\dfrac {2}{x}##+##\dfrac {6}{x}##
##y##=##\dfrac {8}{x}##
i hope am getting it right...

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chwala said:
Homework Statement:: Solve ##x\dfrac {dy}{dx}+y##=##x^{-1/2}##, given that ##y=2## when ##x=4##
Relevant Equations:: First order ode

From my working...I am getting,
##xy=####\int x^{-1/2}\ dx##
##y##=##\dfrac {2}{x}##+##\dfrac {k}{x}##
How did you get the last line in the above quote?

Mark44
George Jones said:
How did you get the last line in the above quote?
think slight error there on my typo...
##y##=##\dfrac {2}{\sqrt x}##+##\dfrac {k}{x}##
applying initial condition, ##y(4)=2##, we get;
##y##=##\dfrac {2}{\sqrt x}##+##\dfrac {4}{ x}##

chwala said:
think slight error there on my typo...
##y##=##\dfrac {2}{\sqrt x}##+##\dfrac {k}{x}##
applying initial condition, ##y(4)=2##, we get;
##y##=##\dfrac {2}{\sqrt x}##+##\dfrac {4}{ x}##
Looks good.

From post #1:
chwala said:
##y##=##\dfrac {2}{x}##+##\dfrac {6}{x}##
##y##=##\dfrac {8}{x}##

i hope am getting it right...
No need to hope for anything -- once you get a solution, you should always check that it satisfies the differential equation. That would have showed that you made an error.

There's an important first step that you didn't show.
##x\dfrac {dy}{dx}+y = x^{-1/2}, x>0##
##\Rightarrow d(xy) = x^{-1/2}##
##\Rightarrow xy = \int x^{-1/2}dx = 2x^{1/2} + k##
and so on.
The second line above takes advantage of the fact that ##\frac {dy}{dx} + y## is the differential of ##xy##.

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Mark44 said:
Looks good.

From post #1:

No need to hope for anything -- once you get a solution, you should always check that it satisfies the differential equation. That would have showed that you made an error.

There's an important first step that you didn't show.
##x\dfrac {dy}{dx}+y = x^{-1/2}, x>0##
##\Rightarrow d(xy) = x^{-1/2}##
##\Rightarrow xy = \int x^{-1/2}dx = 2x^{1/2} + k##
and so on.
The second line above takes advantage of the fact that ##\frac {dy}{dx} + y## is the differential of ##xy##.
Thanks Mark, I actually have all the steps on paper,I just went ahead and posted final steps...true, I should be able verify the solution by differentiation...cheers

ok let me verify my solution, given
##x\dfrac {dy}{dx}+y##=##x^{-1/2}.##
We shall verify the equation above by having,
##x(-x^{-3/2} - 4x^{-2})+2x^{-1/2}+4x^{-1}##=
##-x^{-0.5} - 4x^{-1} +2x^{-0.5}+4x^{-1} ##= ## x^{-0.5}## bingo!

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