# [SOLVED]3.3.003 Find the equation of the curve that passes through the point (1,2)

#### karush

##### Well-known member
Find the equation of the curve that passes through the point $(1,2)$ and has a slope of $(3+\dfrac{1}{x})y$ at any point $(x,y)$ on the curve.
ok this is weird I woild assume the curve would be an parabola and an IVP soluiton...

#### MarkFL

Staff member
You are essentially being asked to solve the IVP:

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

#### karush

##### Well-known member
$\dfrac{1}{y}dy =\left(3+\dfrac{1}{x}\right) dx + C$

then $\int$ both sides??

#### MarkFL

Staff member
I'd suggest you wait until you integrate to introduce a constant of integration. Or use definite integrals with the boundaries as the limits.

#### karush

##### Well-known member
$\ln y = 3x+\ln x$
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!!

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#### MarkFL

Staff member
You are essentially being asked to solve the IVP:

$$\displaystyle \d{y}{x}=\left(3+\frac{1}{x}\right)y$$ where $$y(1)=2$$.
I would next separate the variables, and in doing so we are dividing by $$y$$, thereby eliminating the trivial solution:

$$\displaystyle y\equiv0$$

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:

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

#### karush

##### Well-known member
so i didn't use the limits properly

wow ... that was a great help
appreciate all the steps

#### skeeter

##### Well-known member
MHB Math Helper
$\ln y = 3x+\ln x$
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!!
$\ln{y} = 3x + \ln{x} + \ln{2} - 3$

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

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

#### karush

##### Well-known member
soi stopped too soon!

have to admit that was an interesting problem

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