More and more interesting: I think I can characterize all first-order linear DE's that have this property.  I'll try to write it up in a blog page.

Very interesting: the Runge-Kutta formulas of order 2, 3 and 4 (I didn't have the formula for order 5 handy) give the exact solution for this DE.  Maybe all RK formulas (for order > 1) do.  

Very interesting: the Runge-Kutta formulas of order 2, 3 and 4 (I didn't have the formula for order 5 handy) give the exact solution for this DE.  Maybe all RK formulas (for order > 1) do.  

This seems to have been added in Maple 8.  See the help page ?updates,Maple8,de

This seems to have been added in Maple 8.  See the help page ?updates,Maple8,de

To add to what Axel said:

Equations involving transcendental functions typically have solutions that can't be expressed in "closed form", so the fact that "some solutions may have been lost" may not be Maple's fault. 
 

We might be able to say more if you posted the actual equation.
 

If you want a numerical solution rather than a symbolic one, you might try fsolve.

By definition, x^p = exp(p ln(x)), where Maple uses the principal branch of ln.

So (x^(2*b))^(1/2) = exp(1/2*ln(x^(2*b))) = exp(1/2*ln(exp(2*b*ln(x))))

Now ln(exp(z)) = z + 2*Pi*I*n where n is an integer such that

-Pi < Im(z) + 2*Pi*n <= Pi

i.e. n = floor((1 - Im(z/Pi))/2)

resulting in exp(1/2*ln(exp(z))) = z if n is even, -z if n is odd. 

In your case,, with z = 2*b*ln(x) = 2*(a+1)*ln(x), 0 < x < 1,, we get n=0 iff
-Pi/(2*ln(x)) > Im(a) >= Pi/(2*ln(x)).  For example, Pi/(2*ln(0.1)) is approximately  -.6821881772. 

> plot(map(Re, [sqrt(0.1^(2+2*I*t)), 0.1^(1+I*t)]), t=-3..3,
 colour=[red,blue]);

By definition, x^p = exp(p ln(x)), where Maple uses the principal branch of ln.

So (x^(2*b))^(1/2) = exp(1/2*ln(x^(2*b))) = exp(1/2*ln(exp(2*b*ln(x))))

Now ln(exp(z)) = z + 2*Pi*I*n where n is an integer such that

-Pi < Im(z) + 2*Pi*n <= Pi

i.e. n = floor((1 - Im(z/Pi))/2)

resulting in exp(1/2*ln(exp(z))) = z if n is even, -z if n is odd. 

In your case,, with z = 2*b*ln(x) = 2*(a+1)*ln(x), 0 < x < 1,, we get n=0 iff
-Pi/(2*ln(x)) > Im(a) >= Pi/(2*ln(x)).  For example, Pi/(2*ln(0.1)) is approximately  -.6821881772. 

> plot(map(Re, [sqrt(0.1^(2+2*I*t)), 0.1^(1+I*t)]), t=-3..3,
 colour=[red,blue]);

One issue in this plot is that the plot driver in Classic seems to use
single-precision real arithmetic.  A work-around is to use

> plotsetup(gdi);

However, even there you can get numerical "staircase" effects if you are determined enough.  Try

> plot(1-10*x^5,x=-0.0008 .. 0.001);

The way to do it correctly with dchange is to include the option known, e.g.:

> Int(F(x), x): 
'%'= PDEtools[dchange](x=phi(t),%,[t], known = [F]);

 Int(F(x),x) = Int(F(phi(t))*diff(phi(t),t),t)

In every version of Maple I've tried, Axel's code gives

n*B^2+sum(A^2*x[i]^2+2*A*x[i]*B,i = 1 .. n)

In every version of Maple I've tried, Axel's code gives

n*B^2+sum(A^2*x[i]^2+2*A*x[i]*B,i = 1 .. n)

GeneralSolution is in the form {C[10](z) = ..., C[51](z) = ...}, the right sides being expressions in z.  I wanted functions rather than expressions: {C[10] = (z -> ...), C[51] = (z -> ...)}, because the boundary conditions involved s and h rather than z.  So for each equation, I replaced the left side by its "0'th operand" which is the name of the function (thus replacing C[10](z) by C[10]) and made the right side into a function using unapply.

GeneralSolution is in the form {C[10](z) = ..., C[51](z) = ...}, the right sides being expressions in z.  I wanted functions rather than expressions: {C[10] = (z -> ...), C[51] = (z -> ...)}, because the boundary conditions involved s and h rather than z.  So for each equation, I replaced the left side by its "0'th operand" which is the name of the function (thus replacing C[10](z) by C[10]) and made the right side into a function using unapply.

I've found the Postscript output in Classic to be better than Standard in many cases.
I think the Classic driver is basically the same as the one used in previous versions of Maple.