Differential Equation, Substitution/Homogeneous

[Destroxia]

Homework Statement

Solve the differential equation:

Here is the books answer to the problem:

Homework Equations

This is a substitution/homogeneous first order differential equation, which can be converted into a separable differential equation.

The Attempt at a Solution

Where am I going wrong? (ignore the paragraph breakers in my work, sorry)

 

[BiGyElLoWhAt]

The expression you have 4 lines from the bottom $\int\frac{-1}{5m}dV$ is rather meaningless. Rather than making the m substitution, try using a trigonometric substitution.

 

[BiGyElLoWhAt]

In general, in order for a substitution to be useful, you need to be able to replace all of the old variables with the new variable(s) to get an integral in terms of the new (including the differential)

 

[Destroxia]

The expression you have 4 lines from the bottom $\int\frac{-1}{5m}dV$ is rather meaningless. Rather than making the m substitution, try using a trigonometric substitution.

How would I go about the trigonometric substitution without the radical? Or can you do it without the radical?

 

[Mark44]

The expression you have 4 lines from the bottom $\int\frac{-1}{5m}dV$ is rather meaningless. Rather than making the m substitution, try using a trigonometric substitution.

The integral in question here, $\int \frac{2V dV}{1 - 5V^2}$ can be done without resorting to a trig substition. As it sits, it's pretty close to $\int \frac{du}{u}$.

 

[Destroxia]

The integral in question here, $\int \frac{2V dV}{1 - 5V^2}$ can be done without resorting to a trig substition. As it sits, it's pretty close to $\int \frac{du}{u}$.

Yes, I just used m instead of u, as the derivative of the bottom cancels out all the V terms in the equation, and leaves me just with ms to integrate. I don't understand what I have done wrong.

 

[BiGyElLoWhAt]

But you haven't gotten rid of all the v's... Ok, yea you did, you just didn't replace the dV with dm. Sorry about that. upon further inspection, you actually have the same answer. At least on the second to last line. Try raising $e^{\text{Left Hand Side}}=e^{\text{Right Hand Side}}$

It's a matter of simplification, and I actually had to go through and do it to make sure it was the same. Thanks @Mark44 for straightening me out

 

[Destroxia]

But you haven't gotten rid of all the v's... Ok, yea you did, you just didn't replace the dV with dm. Sorry about that. upon further inspection, you actually have the same answer. At least on the second to last line. Try raising $e^{\text{Left Hand Side}}=e^{\text{Right Hand Side}}$

It's a matter of simplification, and I actually had to go through and do it to make sure it was the same. Thanks @Mark44 for straightening me out

Sorry! I forgot to replace the dV, you are right! I tried raising it to the e, earlier in the day but I suppose I didn't simplify correctly... Thank you for all your help!

 

[Destroxia]

The integral in question here, $\int \frac{2V dV}{1 - 5V^2}$ can be done without resorting to a trig substition. As it sits, it's pretty close to $\int \frac{du}{u}$.

But you haven't gotten rid of all the v's... Ok, yea you did, you just didn't replace the dV with dm. Sorry about that. upon further inspection, you actually have the same answer. At least on the second to last line. Try raising $e^{\text{Left Hand Side}}=e^{\text{Right Hand Side}}$

It's a matter of simplification, and I actually had to go through and do it to make sure it was the same. Thanks @Mark44 for straightening me out

Sorry to re-awaken this problem, but I have gone through the steps of simplifying it as follows...

I get this answer instead of the answer to the problem provided up top.

 

[BiGyElLoWhAt]

No problem.
$e^{2*ln|x|}=e^{ln|x^2|}$
You have a -1/5 exponent that didn't get handled properly.

 

[SammyS]

Sorry to re-awaken this problem, but I have gone through the steps of simplifying it as follows...

I get this answer instead of the answer to the problem provided up top.

Your initial result looked better. You did have some errors in the steps, but the result appears to agree with what the book has.

 

[BiGyElLoWhAt]

Alternatively, between the 3rd and 4th lines, you put the x^5 in the denominator as if it were negative, but its positive. It should stay in the numerator. I see what you did with the 1/5 now.

 

[Destroxia]

Alternatively, between the 3rd and 4th lines, you put the x^5 in the denominator as if it were negative, but its positive. It should stay in the numerator. I see what you did with the 1/5 now.

Yeah, I just figured that out and submitted my work. I was so intent on seeing it come out the way the book had it, I completely ignored the logarithmic rules. Thank you for your time on this problem!

 

[BiGyElLoWhAt]

No problemo señor