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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