Solving a Linear Equation with Initial Condition: $2xy'+y=6x$, $x>0$, $y(4)=20$

[ineedhelpnow]

Hello (Nod)
Can someone help get me started on this problem? Like a setup or direct on how to set it up.

$2xy'+y=6x$, $x>0$, $y(4)=20$

 

[MarkFL]

The first thing I would do is express the ODE in the standard linear form:

\(\displaystyle \d{y}{x}+P(x)y=Q(x)\)

Next, you will want to calculate your integrating factor $\mu(x)$...:D

 

[ineedhelpnow]

$y'+\frac{1}{2x}y=3$
$I(x)=e^{\int \ \frac{1}{2x}dx}=e^{\frac{\ln\left({2x}\right)}{2}}$
$e^{\frac{\ln\left({2x}\right)}{2}}*y'+\frac{y}{2x}*e^{\frac{\ln\left({2x}\right)}{2}}=3e^{\frac{\ln\left({2x}\right)}{2}}$is that right?

 

[MarkFL]

I would simplify your integrating factor using the properties of exponents and logs:

\(\displaystyle a^{b\log_a(c)}=a^{\log_a\left(c^b\right)}=c^b\)

 

[ineedhelpnow]

how do i use the y(4)=20?

 

[MarkFL]

how do i use the y(4)=20?

You use that (the initial conditions) after you have obtained the general solution to the ODE, to solve for the constant of integration.

 

[Ackbach]

Another way to approach this equation is to recognize that it is Cauchy-Euler. The substitution $x=e^{u}$ makes it linear with constant coefficients.

 

[Prove It]

$y'+\frac{1}{2x}y=3$
$I(x)=e^{\int \ \frac{1}{2x}dx}=e^{\frac{\ln\left({2x}\right)}{2}}$
$e^{\frac{\ln\left({2x}\right)}{2}}*y'+\frac{y}{2x}*e^{\frac{\ln\left({2x}\right)}{2}}=3e^{\frac{\ln\left({2x}\right)}{2}}$is that right?

Your integrating factor is wrong.

$\displaystyle \begin{align*} \int{ \frac{1}{2x} \, \mathrm{d}x } &= \frac{1}{2} \int{ \frac{1}{x} \,\mathrm{d}x} \\ &= \frac{1}{2}\ln{ \left| x \right| } + C \end{align*}$

so that means an acceptable integrating factor is

$\displaystyle \begin{align*} \mathrm{e}^{\int{ \frac{1}{2x}\,\mathrm{d}x} } &= \mathrm{e}^{ \frac{1}{2} \ln{ (x) } } \\ &= \mathrm{e}^{ \ln{ \left[ x^{\frac{1}{2}} \right] } } \\ &= x^{\frac{1}{2}} \end{align*}$

 

[ineedhelpnow]

Linear Equation cont

did i do up to here right?

$2xy'+y=6x$

$y'+\frac{1}{2x}y=3$

$I(x)=e^{\int \ \frac{1}{2x}dx}=e^{\frac{1}{2} \ln\left({x}\right)}=e^{\ln\left({x^{1/2}}\right)}=x^{1/2}$

$y'\sqrt{x}+\frac{\sqrt{x}}{2x}*y=3\sqrt{x}$

$y'\sqrt{x}+\frac{y}{2 \sqrt{x}}*y=3\sqrt{x}$

$(y \sqrt{x})'=3\sqrt{x}$

it looks like i made a mistake but i don't know where

 

[Prove It]

did i do up to here right?

$2xy'+y=6x$

$y'+\frac{1}{2x}y=3$

$I(x)=e^{\int \ \frac{1}{2x}dx}=e^{\frac{1}{2} \ln\left({x}\right)}=e^{\ln\left({x^{1/2}}\right)}=x^{1/2}$

$y'\sqrt{x}+\frac{\sqrt{x}}{2x}*y=3\sqrt{x}$

$y'\sqrt{x}+\frac{y}{2 \sqrt{x}}*y=3\sqrt{x}$

$(y \sqrt{x})'=3\sqrt{x}$

it looks like i made a mistake but i don't know where

It's all correct so far :)

 

[ineedhelpnow]

really? so all i do now is integrate $3\sqrt{x}$ and divide it by $\sqrt{x}$

 

[Prove It]

really? so all i do now is integrate $3\sqrt{x}$ and divide it by $\sqrt{x}$

Yes, don't forget your integration constant.

 

[Prove It]

$y'\sqrt{x}+\frac{y}{2 \sqrt{x}}*y=3\sqrt{x}$

Actually there is a small error here, it should be $\displaystyle \begin{align*} y' \,\sqrt{x} + \frac{1}{2\sqrt{x}}\,y = 3\sqrt{x} \end{align*}$, but since you came to the right conclusion I expect this is just a typo.

 

[ineedhelpnow]

yep, it was a typo. i didnt meant to put the $*y$

 

[ineedhelpnow]

so the original asks to find the solution of the initial value problem and it also gives me y(4)=20.

$y \sqrt{x}=\int \ 3\sqrt{x}dx$

$y \sqrt{x}=2x^{3/2}$

$y=2x+C$

$20=2(4)+C$
$20=8+C$
$C=12$

$y=2x+12$

(Yes), (No), maybe so?

 

[Prove It]

so the original asks to find the solution of the initial value problem and it also gives me y(4)=20.

$y \sqrt{x}=\int \ 3\sqrt{x}dx$

$y \sqrt{x}=2x^{3/2}$

$y=2x+C$

$20=2(4)+C$
$20=8+C$
$C=12$

$y=2x+12$

(Yes), (No), maybe so?

No, the moment you integrate is when you bring in your integration constant!

$\displaystyle \begin{align*} y\,\sqrt{x} &= 2x^{\frac{3}{2}} + C \\ y &= 2x + \frac{C}{\sqrt{x}} \end{align*}$

NOW substitute your point.

 

[MarkFL]

I have merged the two threads pertaining to the same question.

If you have a new question, then you should start a new thread, but if you are continuing your work on an existing question, then you should continue your posts in that thread. This way, we don't have multiple threads dealing with the same question.

 

[ineedhelpnow]

$y=2x+\frac{C}{\sqrt{x}}$

$20=2(4)+\frac{C}{\sqrt{4}}$

$20=8+\frac{C}{2}$

$\frac{C}{2}=12$

$C=24$

$y=2x+\frac{24}{\sqrt{x}}$

 

[Prove It]

$y=2x+\frac{C}{\sqrt{x}}$

$20=2(4)+\frac{C}{\sqrt{4}}$

$20=8+\frac{C}{2}$

$\frac{C}{2}=12$

$C=24$

$y=2x+\frac{24}{\sqrt{x}}$

That is correct, well done :)