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#### karush

##### Well-known member

- Jan 31, 2012

- 2,678

ok this is weird I woild assume the curve would be an parabola and an IVP soluiton...

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- Thread starter
- #1

- Jan 31, 2012

- 2,678

ok this is weird I woild assume the curve would be an parabola and an IVP soluiton...

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- #3

- Jan 31, 2012

- 2,678

$\dfrac{1}{y}dy =\left(3+\dfrac{1}{x}\right) dx + C$

then $\int$ both sides??

then $\int$ both sides??

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- #5

- Jan 31, 2012

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$\ln y = 3x+\ln x$

ok so we got $y(1)=2$, $\quad x=1\quad y=2$

$\ln 2 = 3(1) + \ln 1 + C$

so

$\ln 2-3 =C$

ummmmm!!

ok so we got $y(1)=2$, $\quad x=1\quad y=2$

$\ln 2 = 3(1) + \ln 1 + C$

so

$\ln 2-3 =C$

ummmmm!!

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- #6

I would next separate the variables, and in doing so we are dividing by \(y\), thereby eliminating the trivial solution:You are essentially being asked to solve the IVP:

\(\displaystyle \d{y}{x}=\left(3+\frac{1}{x}\right)y\) where \(y(1)=2\).

\(\displaystyle y\equiv0\)

And so we have:

\(\displaystyle \frac{1}{y}\,dy=\left(3+\frac{1}{x}\right)\,dx\)

Integrate, using the boundaries as limits:

\(\displaystyle \int_2^y \frac{1}{u}\,du=\int_1^x 3+\frac{1}{v}\,dv\)

\(\displaystyle \left[\ln|u|\right]_2^y=\left[3v+\ln|v|\right]_1^x\)

\(\displaystyle \ln|y|-\ln(2)=(3x+\ln|x|)-(3+\ln(1))\)

\(\displaystyle \ln|y|=3x+\ln|x|-3+\ln(2)\)

This implies:

\(\displaystyle y(x)=2xe^{3(x-1)}\)

We could also have written the ODE as:

\(\displaystyle \d{y}{x}-\left(3+\frac{1}{x}\right)y=0\)

Compute the integrating factor:

\(\displaystyle \mu(x)=\exp\left(-\int 3+\frac{1}{x}\,dx\right)=\frac{e^{-3x}}{x}\)

And the ODE becomes:

\(\displaystyle \frac{e^{-3x}}{x}\d{y}{x}-\frac{e^{-3x}}{x}\left(3+\frac{1}{x}\right)y=0\)

\(\displaystyle \frac{d}{dx}\left(\frac{e^{-3x}}{x}y\right)=0\)

\(\displaystyle \frac{e^{-3x}}{x}y=c_1\)

\(\displaystyle y(x)=c_1xe^{3x}\)

\(\displaystyle y(1)=c_1e^3=2\implies c_1=2e^{-3}\)

Hence:

\(\displaystyle y(x)=2xe^{3(x-1)}\)

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- #7

- Jan 31, 2012

- 2,678

so i didn't use the limits properly

wow ... that was a great help

appreciate all the steps

wow ... that was a great help

appreciate all the steps

- Mar 1, 2012

- 655

$\ln{y} = 3x + \ln{x} + \ln{2} - 3$$\ln y = 3x+\ln x$

ok so we got $y(1)=2$, $\quad x=1\quad y=2$

$\ln 2 = 3(1) + \ln 1 + C$

so

$\ln 2-3 =C$

ummmmm!!

$\ln{y} = 3(x-1) + \ln(2x)$

$y = 2x \cdot e^{3(x-1)}$

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- #9

- Jan 31, 2012

- 2,678

soi stopped too soon!

have to admit that was an interesting problem

have to admit that was an interesting problem

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