Answer: [tex]\displaystyle y = \frac{1}{9}(3x-1)^{6}+\frac{8}{9}\\\\[/tex]
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Explanation:
Multiply dx on each side to move it over.
Doing so gets us to
[tex]dy = 2(kx-1)^5dx[/tex]
Next, integrate both sides with respect to the dy and dx terms
[tex]dy = 2(kx-1)^5dx\\\\
\displaystyle \int dy = \int 2(kx-1)^5dx\\\\
\displaystyle y = 2\int (kx-1)^5dx\\\\[/tex]
We can do a u-substitution here.
Let u = kx-1 which leads to du/dx = k which turns into [tex]du = kdx[/tex] and rearranges to [tex]dx = \frac{du}{k}[/tex]
So the steps continuing forward could look like
[tex]\displaystyle y = 2\int (kx-1)^5dx\\\\
\displaystyle y = 2\int u^5*\frac{du}{k}\\\\
\displaystyle y = \frac{2}{k}\int u^5du\\\\
\displaystyle y = \frac{2}{k}\left(\frac{1}{5+1}u^{5+1}+C\right)\\\\
\displaystyle y = \frac{2}{k}\left(\frac{1}{6}u^{6}+C\right)\\\\
\displaystyle y = \frac{2}{k}\left(\frac{1}{6}(kx-1)^{6}+C\right)\\\\[/tex]
Let's plug in (x,y) = (0,1) and do a bit of algebra and solve for C. We'll get some equation in terms of k on the right hand side
[tex]\displaystyle y = \frac{2}{k}\left(\frac{1}{6}(kx-1)^{6}+C\right)\\\\
\displaystyle 1 = \frac{2}{k}\left(\frac{1}{6}(k*0-1)^{6}+C\right)\\\\
\displaystyle 1 = \frac{2}{k}\left(\frac{1}{6}+C\right)\\\\
\displaystyle k = 2\left(\frac{1}{6}+C\right)\\\\
k = \frac{1}{3}+2C[/tex]
[tex]3k = 1+6C\\\\
6C = 3k-1\\\\
C = \frac{3k-1}{6}\\\\[/tex]
Now let's plug in (x,y) = (1,8) along with that C value. Then solve for k to get an actual number.
[tex]\displaystyle y = \frac{2}{k}\left(\frac{1}{6}(kx-1)^{6}+C\right)\\\\
8 = \frac{2}{k}\left(\frac{1}{6}(k*1-1)^{6}+\frac{3k-1}{6}\right)\\\\[/tex]
I'm skipping a bunch of steps, but you should get to [tex]k = 3[/tex] when solving that equation. You can use a graphing calculator to find the root of that function (think of k as the input x) and the x intercept is 3.
This leads to,
[tex]C = \frac{3k-1}{6}\\\\
C = \frac{3*3-1}{6}\\\\
C = \frac{8}{6}\\\\
C = \frac{4}{3}\\\\[/tex]
Therefore,
[tex]\displaystyle y = \frac{2}{k}\left(\frac{1}{6}(kx-1)^{6}+C\right)\\\\
y = \frac{2}{3}\left(\frac{1}{6}(3x-1)^{6}+\frac{4}{3}\right)\\\\
\boldsymbol{y = \frac{1}{9}(3x-1)^{6}+\frac{8}{9}} \ \textbf{ which is the final answer}\\\\[/tex]
If k = 3, then,
[tex]\displaystyle \frac{dy}{dx} = 2(kx-1)^5\\\\
\displaystyle \frac{dy}{dx} = 2(3x-1)^5\\\\[/tex]
If you were to differentiate the answer function with respect to x, then you should get the dy/dx expression mentioned.
