Seperation of Variables Problem

[Jason76]

Is this right? How would you graph the general solution?

$\displaystyle \dfrac{dy}{dx} = \dfrac{y + 1}{6x - 9}$

$\displaystyle \dfrac{dy}{dx}(dx) = \dfrac{y + 1}{6x - 9}(dx)$

$\displaystyle dy = \dfrac{y + 1}{6x - 9} dx$

$\displaystyle (\dfrac{1}{y + 1}) dy = \dfrac{y + 1}{6x - 9}(\dfrac{1}{y + 1})dx$

$\displaystyle \dfrac{1}{y + 1} dy = \dfrac{1}{6x - 9} dx$

$\displaystyle \int \dfrac{1}{y + 1} dy = \int \dfrac{1}{6x + 9} dx$

$\displaystyle \ln (y + 1) + C_{a} = \dfrac{1}{6}\ln (6x + 9) + C_{b}$

$\displaystyle \ln (y + 1) + C_{a} - C_{a} = \dfrac{1}{6}\ln (6x + 9) + C_{b} - C_{a}$

$\displaystyle \ln (y + 1) =\dfrac{1}{6}\ln (6x + 9) + C$

$\displaystyle e^{\ln (y + 1)} = e^{\frac{1}{6}\ln (6x + 9) + C}$

$\displaystyle y + 1 = \dfrac{1}{6}(6x + 9)(e^{c})$

$\displaystyle e^{c} = c_{1}$

$\displaystyle y + 1 = \dfrac{1}{6}(6x + 9)(c_{1})$

$\displaystyle y + 1 = x + \dfrac{3}{2}(c_{1})$

$\displaystyle y + 1 - 1 = x + \dfrac{3}{2}(c_{1}) - 1$

$\displaystyle y = x + \dfrac{3}{2} (c_{1}) - 1$ is the general solution.

 

[soroban]

Hello, Jason76!

You dropped a "6" . . .

$\displaystyle \dfrac{dy}{dx} \:=\: \dfrac{y + 1}{6x - 3}$

We have: .$\displaystyle \dfrac{dy}{dx} \;=\;\dfrac{y+1}{3(2x-1)}$

Separate variables: .$\displaystyle \dfrac{3\,dy}{y+1} \;=\;\dfrac{dx}{2x-1}$

Integrate: .$\displaystyle \displaystyle 3\int\frac{dy}{y+1} \;=\;\int \frac{dx}{2x-1}$

. . . . . . . . $\displaystyle 3\ln|y+1| \;=\;\frac{1}{2}\ln|2x-1| + C_1$

. . . . . . . . . $\displaystyle \ln|y+1| \;=\;\frac{1}{6}\ln|2x -1| + C_2$

. . . . . . . . . $\displaystyle \ln|y+1| \;=\;\ln(2x-1)^{\frac{1}{6}} + \ln C$

. . . . . . . . . $\displaystyle \ln|y+1| \;=\;\ln\left|C(2x-1)^{\frac{1}{6}}\right|$

. . . . . . . . . . . .$\displaystyle y+1 \;=\;C(2x-1)^{\frac{1}{6}}$

n . . . . . . . . . . . . . $\displaystyle y \;=\;C(2x-1)^{\frac{1}{6}}-1$

 

[Jason76]

In response to Soroban:

Typo, the 3s in the equations near the top should have been 9. I went back and edited the post.

Anyhow, even if the number was 3, it seems like you could use the distributive property right off. That would turn the equation into (6x - 9), and then you could solve it my way.

 

[Deleted member 4993]

Is this right? How would you graph the general solution?

$\displaystyle \dfrac{dy}{dx} = \dfrac{y + 1}{6x - 9}$

$\displaystyle \dfrac{dy}{dx}(dx) = \dfrac{y + 1}{6x - 9}(dx)$

$\displaystyle dy = \dfrac{y + 1}{6x - 9} dx$

$\displaystyle (\dfrac{1}{y + 1}) dy = \dfrac{y + 1}{6x - 9}(\dfrac{1}{y + 1})dx$

$\displaystyle \dfrac{1}{y + 1} dy = \dfrac{1}{6x - 9} dx$

$\displaystyle \int \dfrac{1}{y + 1} dy = \int \dfrac{1}{6x + 9} dx$

$\displaystyle \ln (y + 1) + C_{a} = \ln (6x + 9) + C_{b}$...... Incorrect .... should be $\displaystyle \ln (y + 1) + C_{a} = \dfrac{1}{6}\ln (6x + 9) + C_{b}$

Fix the rest accordingly

$\displaystyle \ln (y + 1) + C_{a} - C_{a} = \ln (6x + 9) + C_{b} - C_{a}$

$\displaystyle \ln (y + 1) = \ln (6x + 9) + C$

$\displaystyle e^{\ln (y + 1)} = e^{\ln (6x + 9) + C}$

$\displaystyle y + 1 = 6x + 9(e^{c})$

$\displaystyle e^{c} = c_{1}$

$\displaystyle y + 1 = 6x + 9(c_{1})$

$\displaystyle y + 1 - 1 = 6x + 9(c_{1}) - 1$

$\displaystyle y = 6x + 9(c_{1}) - 1$ is the general solution.

.

 

[Jason76]

.

Where did the 1/6 come from?

Oh wait, I see, from u substitution. Original post has been edited.

 

[Deleted member 4993]

Is this right? How would you graph the general solution?

$\displaystyle \dfrac{dy}{dx} = \dfrac{y + 1}{6x - 9}$

$\displaystyle \dfrac{dy}{dx}(dx) = \dfrac{y + 1}{6x - 9}(dx)$

$\displaystyle dy = \dfrac{y + 1}{6x - 9} dx$

$\displaystyle (\dfrac{1}{y + 1}) dy = \dfrac{y + 1}{6x - 9}(\dfrac{1}{y + 1})dx$

$\displaystyle \dfrac{1}{y + 1} dy = \dfrac{1}{6x - 9} dx$

$\displaystyle \int \dfrac{1}{y + 1} dy = \int \dfrac{1}{6x + 9} dx$

$\displaystyle \ln (y + 1) + C_{a} = \dfrac{1}{6}\ln (6x + 9) + C_{b}$

$\displaystyle \ln (y + 1) + C_{a} - C_{a} = \dfrac{1}{6}\ln (6x + 9) + C_{b} - C_{a}$

$\displaystyle \ln (y + 1) =\dfrac{1}{6}\ln (6x + 9) + C$

$\displaystyle e^{\ln (y + 1)} = e^{\frac{1}{6}\ln (6x + 9) + C}$ .... incorrect .... should be

ln (y + 1) = ln [C₁*(6x + 9)^(1/6)]

Fix the rest accordingly.....

$\displaystyle y + 1 = \dfrac{1}{6}(6x + 9)(e^{c})$

$\displaystyle e^{c} = c_{1}$

$\displaystyle y + 1 = \dfrac{1}{6}(6x + 9)(c_{1})$

$\displaystyle y + 1 = x + \dfrac{3}{2}(c_{1})$

$\displaystyle y + 1 - 1 = x + \dfrac{3}{2}(c_{1}) - 1$

$\displaystyle y = x + \dfrac{3}{2} (c_{1}) - 1$ is the general solution.

.
You should be able to tell whether your solution is correct (or not) by substituting it back into the original ODE.

 

[HallsofIvy]

Is this right? How would you graph the general solution?

$\displaystyle \dfrac{dy}{dx} = \dfrac{y + 1}{6x - 9}$

$\displaystyle \dfrac{dy}{dx}(dx) = \dfrac{y + 1}{6x - 9}(dx)$

$\displaystyle dy = \dfrac{y + 1}{6x - 9} dx$

$\displaystyle (\dfrac{1}{y + 1}) dy = \dfrac{y + 1}{6x - 9}(\dfrac{1}{y + 1})dx$

$\displaystyle \dfrac{1}{y + 1} dy = \dfrac{1}{6x - 9} dx$

$\displaystyle \int \dfrac{1}{y + 1} dy = \int \dfrac{1}{6x + 9} dx$

$\displaystyle \ln (y + 1) + C_{a} = \dfrac{1}{6}\ln (6x + 9) + C_{b}$

$\displaystyle \ln (y + 1) + C_{a} - C_{a} = \dfrac{1}{6}\ln (6x + 9) + C_{b} - C_{a}$

$\displaystyle \ln (y + 1) =\dfrac{1}{6}\ln (6x + 9) + C$

$\displaystyle e^{\ln (y + 1)} = e^{\frac{1}{6}\ln (6x + 9) + C}$

$\displaystyle y + 1 = \dfrac{1}{6}(6x + 9)(e^{c})$

$\displaystyle e^{c} = c_{1}$

$\displaystyle y + 1 = \dfrac{1}{6}(6x + 9)(c_{1})$

$\displaystyle y + 1 = x + \dfrac{3}{2}(c_{1})$

You have dropped the parentheses needed here: you should have
[itex]y+ 1= (x+ dfrac{3}{2}(c_1)[itex]

$\displaystyle y + 1 - 1 = x + \dfrac{3}{2}(c_{1}) - 1$

$\displaystyle y = x + \dfrac{3}{2} (c_{1}) - 1$ is the general solution.

No. y= (x+ 3/2)c_1- 1.