Question

3r = (dr/dtheta) - theta^3. Determine if it's linear, seperable, neither, or both

Answer 1

\[3r=\frac{dr}{d\theta}-\theta^3\]

Answer 2

can we get all the r,dr's on one side and theta,dthetas on the other?

Answer 3

So this is
\[3r=\frac{dr}{d\theta}-\theta^3\]
I will replace the substitution:
\(r=y\)
\(\theta=x\)
Now we have:
\[3y=\frac{dy}{dx}-x^3\]
We can re-arrange this to:
\[\frac{dy}{dx}-3y=x^3\]
This is in the form of:
\[\frac{dy}{dx}+p(x)y=q(x)\]
So therefore, it is linear and we can separate it

Answer 4

We isolate differential equations of the form:
\[\frac{dy}{dx}+p(x)y=q(x)\]
By using this formula:
\[y=\frac{\int{e^{p(x)dx}q(x)}+C}{e^{p(x) dx}}\]

Answer 5

This makes so much more since.

Answer 6

Or simplified:
\[y=\frac{\int{\mu(x)q(x)}+C}{\mu(x)}\] Where \(\mu(x)=e^{\int{p(x)dx}}\)
And I made a mistake on the above one

Answer 7

So we can simplify this by stating:
\[\eqalign{
&p(x)=-3 \\
&q(x)=x^3 \\
}\]

So:
\[y=\frac{\int{\mu(x)x^3}+c_1}{\mu(x)}\]
Where:
\[\mu(x)=e^{\int{-3 dx}}=e^{-3x}\]
So therefore:
\[y=\frac{\int{e^{-3x}x^3}+c_1}{e^{-3x}}\]

Answer 8

Very thorough explanation. Thank you

Answer 9

Your welcome :) I could go a step further and simplify evaluating the integrals if you want. Or I can stop there if you're happy enough.

Answer 10

This is fine although I do need help with another question if you're interested.