# Differential Equation, Substitution/Homogeneous

• Destroxia
In summary, the book suggests that you try a trigonometric substitution instead of a m substitution for the equation in question.
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)

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.

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)

BiGyElLoWhAt said:
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?

BiGyElLoWhAt said:
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}##.

Mark44 said:
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.

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

BiGyElLoWhAt said:
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!

Mark44 said:
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}##.

BiGyElLoWhAt said:
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...

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

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

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

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.

BiGyElLoWhAt said:
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!

No problemo señor

## 1. What is a differential equation?

A differential equation is a mathematical equation that relates one or more functions and their derivatives. It describes how a quantity changes over time or space, and is often used to model natural phenomena in fields such as physics, engineering, and economics.

## 2. What is substitution in differential equations?

Substitution is a method used to solve certain types of differential equations, particularly those that are linear or separable. It involves replacing one variable with another, often to simplify the equation and make it easier to solve.

## 3. What is a homogeneous differential equation?

A homogeneous differential equation is one in which all terms contain the dependent variable and its derivatives, with no other independent variables. In other words, the equation is "homogeneous" because all terms have the same degree or order.

## 4. How do I solve a homogeneous differential equation using substitution?

To solve a homogeneous differential equation using substitution, you first need to identify the appropriate substitution to make. This is often done by looking for patterns in the equation or by using a known solution as a guide. Then, you substitute the new variable into the original equation and solve for the new variable. Finally, you substitute the original variable back in to get the final solution.

## 5. Can substitution always be used to solve a homogeneous differential equation?

No, substitution is not always applicable to solve a homogeneous differential equation. It typically works best for equations that are linear or separable, and may not work for more complex equations. In those cases, other methods such as variation of parameters or power series may be used.

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