Solve the first order differential equation

[chwala]

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...

 

[George Jones]

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?

 

[chwala]

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;
$y$=$\dfrac {2}{\sqrt x}$+$\dfrac {4}{ x}$

 

[Mark44]

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:

$y$=$\dfrac {2}{x}$+$\dfrac {6}{x}$
$y$=$\dfrac {8}{x}$

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 $xy$.

 

[chwala]

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!