## find particular soln to dy/dx=3-y/x using analytical method

Limits, differentiation, related rates, integration, trig integrals, etc.
BRAIN
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### find particular soln to dy/dx=3-y/x using analytical method

How to find particular solution using analytical method?
if dy/dx=3-y/x, and there is an initial condition that y=2 when x=1

stapel_eliz
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How to find particular solution using analytical method?
if dy/dx=3-y/x....
Does the above mean this?

. . . . .$\frac{dy}{dx}\, =\, 3\, -\, \frac{y}{x}$

Or this?

. . . . .$\frac{dy}{dx}\, =\, \frac{3\, -\, y}{x}$

What does your book mean by "the analytical method"? How far have you gotten in applying this method?

BRAIN
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### Re: find particular soln to dy/dx=3-y/x using analytical method

it is :
. . . . .$\frac{dy}{dx}\, =\, 3\, -\, \frac{y}{x}$

I think Analytical solutions means relatively few different equations can be solved analytically, gives exact solution.. maybe question meant that.
Also I have another question and post here that using the same equation and finding numerical solution using euler method, maybe saying use and analytical method meant to seperate this two questions!

I think I need to use dy/dx + Py = Q but I dont know how to deal with this equation.
Hope this is clear to help.

Thank you.

Martingale
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### Re: find particular soln to dy/dx=3-y/x using analytical method

it is :
. . . . .$\frac{dy}{dx}\, =\, 3\, -\, \frac{y}{x}$

I think Analytical solutions means relatively few different equations can be solved analytically, gives exact solution.. maybe question meant that.
Also I have another question and post here that using the same equation and finding numerical solution using euler method, maybe saying use and analytical method meant to seperate this two questions!

I think I need to use dy/dx + Py = Q but I dont know how to deal with this equation.
Hope this is clear to help.

Thank you.
Use an Integrating Factor.

FrozenBlood
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Joined: Wed Feb 04, 2009 11:31 pm
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### Re: find particular soln to dy/dx=3-y/x using analytical method

Is this for a calculus class or a differential equations class?

Anyway, $\frac{dy}{dx}\, =\, 3\, -\, \frac{y}{x}$ is a pretty straightforward linear first-order ODE.

First, rewrite it in standard form: $\frac{dy}{dx}\, + P(x)y = Q(x)$

In standard form, $\frac{dy}{dx}\, =\, 3\, -\, \frac{y}{x}$ is $\frac{dy}{dx}\, +\, \frac{y}{x}=\, 3\$.

Now get the integrating factor $\mu(x) = e^{\int P(x)dx$:

$\mu(x) = e^{\int \frac{1}{x}dx$
$\mu(x) = e^{ln|x| + C}$
$\mu(x) = Ce^{ln|x|}$
$\mu(x) = Cx$
$\mu(x) = x$ (let C = 1)

Now multiply x through both sides of $\frac{dy}{dx}\, +\, \frac{y}{x}=\, 3\$:
$x\frac{dy}{dx}\, +\, y=\, 3x\$

Rewrite the left side as: $\frac{d}{dx}\(xy) =\, 3x\$

Multiply both sides by dx:

$d(xy) =\, 3xdx$

Now take the integral of both sides:

$\int\ d(xy) = \int\ 3xdx$
$xy = \frac{3}{2}x^{2} + C$
$y = \frac{3}{2}x + Cx^{-1}$

And that's the general solution. But this is an IVP, so you need to solve for C using the initial condition y(1) = 2:

$2 = \frac{3}{2}(1) + C(1)^{-1}$
$2 = \frac{3}{2} + C$
$C = 2 - \frac{3}{2}$
$C = \frac{1}{2}$

So the particular solution to the IVP is $y = \frac{3}{2}x + \frac{1}{2}x^{-1}$