Seperation of Variables Problem

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Jason76

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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.
 
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Hello, Jason76!

You dropped a "6" . . .

\(\displaystyle \dfrac{dy}{dx} \:=\: \dfrac{y + 1}{6x - 3}\)
Click to expand...

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 . . . . . . . . . . . . . \(\displaystyle y \;=\;C(2x-1)^{\frac{1}{6}}-1 \)
 
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.
 
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Jason76 said:
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.
Click to expand...
.
 
Jason76 said:
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.
Click to expand...
.
You should be able to tell whether your solution is correct (or not) by substituting it back into the original ODE.
 
Last edited by a moderator:
Jason76 said:
Is this right? How would you graph the general solution?

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

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

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

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

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

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

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

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

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

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

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

\(\displaystyle e^{c} = c_{1}\)

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

\(\displaystyle y + 1 = x + \dfrac{3}{2}(c_{1})\)
Click to expand...
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.
Click to expand...
No. y= (x+ 3/2)c_1- 1.
 
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