Assistance needed for this ODE problem

[heaviside]

Hi,The problem I am trying to solve is in a section on first order ODEs. It is problem 25 in section 2.1 of Boyce & DiPrima's Elementary Diff Eq (5th Ed). The problem serves as an introduction to the variation of parameters, but again, it is in the first section of the book that introduces first order ODEs.

Find A(x) from (i). Then substitute for A(x) in (ii). Verify that the solution verifies with (iii).

(i)
\(\displaystyle
A'(x) = g(x) \exp \lbrack \int p(x)\,dx \rbrack
\)

(ii)
\(\displaystyle
y = A(x) \exp \lbrack -\int p(x)\,dx \rbrack
\)

(iii)
\(\displaystyle
y = \frac{\int \mu(x)g(x)dx + c}{\mu(x)}
\)

This is what I have so far:

Integrate (by parts) equation (i):
\(\displaystyle
\int udv = uv-\int vdu
\)
Let
\(\displaystyle u = \exp \lbrack \int p(x)\,dx \rbrack \)
\(\displaystyle du = p(x) \exp \lbrack \int p(x)\,dx \rbrack \)
\(\displaystyle v = g(x) \)
\(\displaystyle dv=g'(x) \)

This yields:
\(\displaystyle
A(x)=g(x) \exp \lbrack \int p(x)\,dx \rbrack - \int g(x)p(x) \exp \lbrack \int p(x)\,dx \rbrack dx
\)

Putting this into equation (ii), yields:
\(\displaystyle
y = \lgroup g(x) \exp \lbrack \int p(x)\,dx \rbrack - \int g(x)p(x) \exp \lbrack \int p(x)\,dx \rbrack dx \rgroup \bullet \exp \lbrack - \int p(x)\,dx \rbrack
\)
and thus (iv)
\(\displaystyle
y = g(x) - \lgroup \int g(x)p(x) \exp \lbrack \int p(x)\,dx \rbrack dx \rgroup \bullet \exp \lbrack - \int p(x)\,dx \rbrack
\)

Now, I perform integration by parts again on the term in round brackets above:
Let
\(\displaystyle u = g(x) \)
\(\displaystyle du = g'(x)\)
\(\displaystyle v = \exp \lbrack \int p(x)\,dx \rbrack \)
\(\displaystyle dv = p(x) \exp \lbrack \int p(x)\,dx \rbrack \)

This yields:
\(\displaystyle
g(x)\exp \lbrack \int p(x)\,dx \rbrack - \int g'(x) \exp \lbrack \int p(x)\,dx \rbrack dx
\)

Now (iv) becomes
\(\displaystyle
y = g(x) - \lgroup g(x)\exp \lbrack \int p(x)\,dx \rbrack - \int g'(x) \exp \lbrack \int p(x)\,dx \rbrack dx \rgroup \exp \lbrack - \int p(x)\,dx \rbrack
\)
Simplifying:
\(\displaystyle
y = \frac{\int g'(x)\exp \lbrack \int p(x)\,dx \rbrack dx}{\exp \lbrack \int p(x)\,dx \rbrack}
\)

Now the book defines
\(\displaystyle
\mu(x) = \exp \lbrack \int p(x)\,dx \rbrack
\)

Therefore, what I have is:
\(\displaystyle
y=\frac{\int \mu(x)g'(x)dx + c}{\mu(x)}
\)

Compare this to (iii) - you will see I have \(\displaystyle g'(x)\) instead of \(\displaystyle g(x)\).

So my question is: where did I go wrong. I've been looking over this for hours.

Thanks,
heaviside

 

[Ackbach]

Re: Assistance needed for this ODE problame

Hi,The problem I am trying to solve is in a section on first order ODEs. It is problem 25 in section 2.1 of Boyce & DiPrima's Elementary Diff Eq (5th Ed). The problem serves as an introduction to the variation of parameters, but again, it is in the first section of the book that introduces first order ODEs.

Find A(x) from (i). Then substitute for A(x) in (ii). Verify that the solution verifies with (iii).

(i)
\(\displaystyle
A'(x) = g(x) \exp \lbrack \int p(x)\,dx \rbrack
\)

(ii)
\(\displaystyle
y = A(x) \exp \lbrack -\int p(x)\,dx \rbrack
\)

(iii)
\(\displaystyle
y = \frac{\int \mu(x)g(x)dx + c}{\mu(x)}
\)

This is what I have so far:

Integrate (by parts) equation (i):
\(\displaystyle
\int udv = uv-\int vdu
\)
Let
\(\displaystyle u = \exp \lbrack \int p(x)\,dx \rbrack \)
\(\displaystyle du = p(x) \exp \lbrack \int p(x)\,dx \rbrack \)
\(\displaystyle v = g(x) \)
\(\displaystyle dv=g'(x) \)

I'm assuming you meant $dv = g'(x) \, dx$. Remember the Golden Rule of Differentials: if you have a differential on one side of an equation, you must have a differential on the other side as well.

This is not what you have in your integral. For by-parts, the integral you start with has to be the $\displaystyle \int u \, dv$. So you're saying, by setting up the by-parts this way, that your original integral is $\displaystyle \int g'(x) e^{ \int p(x) \, dx} \, dx$. But it isn't. Your original integral is $\displaystyle \int g(x) e^{ \int p(x) \, dx} \, dx$.

Try propagating this correction through your derivations, and see if it comes out right.

This yields:
\(\displaystyle
A(x)=g(x) \exp \lbrack \int p(x)\,dx \rbrack - \int g(x)p(x) \exp \lbrack \int p(x)\,dx \rbrack dx
\)

Putting this into equation (ii), yields:
\(\displaystyle
y = \lgroup g(x) \exp \lbrack \int p(x)\,dx \rbrack - \int g(x)p(x) \exp \lbrack \int p(x)\,dx \rbrack dx \rgroup \bullet \exp \lbrack - \int p(x)\,dx \rbrack
\)
and thus (iv)
\(\displaystyle
y = g(x) - \lgroup \int g(x)p(x) \exp \lbrack \int p(x)\,dx \rbrack dx \rgroup \bullet \exp \lbrack - \int p(x)\,dx \rbrack
\)

Now, I perform integration by parts again on the term in round brackets above:
Let
\(\displaystyle u = g(x) \)
\(\displaystyle du = g'(x)\)
\(\displaystyle v = \exp \lbrack \int p(x)\,dx \rbrack \)
\(\displaystyle dv = p(x) \exp \lbrack \int p(x)\,dx \rbrack \)

This yields:
\(\displaystyle
g(x)\exp \lbrack \int p(x)\,dx \rbrack - \int g'(x) \exp \lbrack \int p(x)\,dx \rbrack dx
\)

Now (iv) becomes
\(\displaystyle
y = g(x) - \lgroup g(x)\exp \lbrack \int p(x)\,dx \rbrack - \int g'(x) \exp \lbrack \int p(x)\,dx \rbrack dx \rgroup \exp \lbrack - \int p(x)\,dx \rbrack
\)
Simplifying:
\(\displaystyle
y = \frac{\int g'(x)\exp \lbrack \int p(x)\,dx \rbrack dx}{\exp \lbrack \int p(x)\,dx \rbrack}
\)

Now the book defines
\(\displaystyle
\mu(x) = \exp \lbrack \int p(x)\,dx \rbrack
\)

Therefore, what I have is:
\(\displaystyle
y=\frac{\int \mu(x)g'(x)dx + c}{\mu(x)}
\)

Compare this to (iii) - you will see I have \(\displaystyle g'(x)\) instead of \(\displaystyle g(x)\).

So my question is: where did I go wrong. I've been looking over this for hours.

Thanks,
heaviside

 

[heaviside]

That's it!

I just went through it again, with the change you suggested, and with more care. And the solution came out.

Thank you Ackbach!

 

[Ackbach]

That's it!

I just went through it again, with the change you suggested, and with more care. And the solution came out.

Thank you Ackbach!

You're very welcome! Have a good one.

 

[blue_raver22]

Hello heaviside,

Thank you for reaching out for assistance with this ODE problem. From the steps that you have provided, it seems like you have approached the problem correctly. However, there may be a small error in the integration by parts that you have performed in the first step.

When integrating by parts, it is important to remember that the integral of a product is not simply the product of the integrals. In your first integration by parts, you have written:

\int udv = uv - \int vdu

However, the correct formula is:

\int udv = uv - \int vdu + C

Where C is the constant of integration. This may be the cause of the discrepancy between your answer and the one provided in the book.

I would suggest going through the integration by parts again, keeping in mind the constant of integration, and see if that leads to the correct solution. If you are still having trouble, you can also try using a different method such as substitution or the variation of parameters to solve the problem.

I hope this helps and good luck with your problem-solving!