Cfrgtkky

The following problem is from the book "Introduction to Ordinary Differential Equations" by
Shepley L. Ross. Below is my solution to the problem, which I believe is wrong. In any case,
it does not match the answer in the book. I am hoping somebody here can tell me where I went wrong.

Thanks,

Bob

Problem:

Given that $y = e^{2x}$ is a solution of
$$ (2x+1)y'' - 4(x+1)y' + 4y = 0$$
find a linearly independent solution by reducing the order. Write the general solution.

Answer:

Let $f(x)$ represent the solution we have.
begin{eqnarray*}
f(x) &=& e^{2x} \
y &=& f(x) v = e^{2x}v \
y' &=& e^{2x} v' + 2ve^{2x} \
y'' &=& e^{2x} v'' + 2e^{2x}v' + 4ve^{2x} + 2v'e^{2x} \
y'' &=& e^{2x} v'' + 4e^{2x}v' + 4ve^{2x} \
end{eqnarray*}
begin{eqnarray*}
(2x+1)(e^{2x} v'' + 4e^{2x}v' + 4ve^{2x}) - 4(x+1)(e^{2x} v' + 2ve^{2x}) + 4e^{2x}v &=& 0 \
(2x+1)( v'' + 4v' + 4v) - 4(x+1)( v' + 2v) + 4v &=& 0 \
(2x+1)( v'' + 4v' + 4v) - 4(x+1)( v') - 4(x+1)(2v) + 4v &=& 0 \
(2x+1)( v'' + 4v' + 4v) - 4(x+1)( v') - 8xv - 8v + 4v &=& 0 \
(2x+1)( v'' + 4v' + 4v) - 4(x+1)( v') - 8xv - 4v &=& 0 \
(2x+1)(v'' + 4v' + 4v) + (2x+1)(4v)- 4(x+1)( v') - 8xv - 4v &=& 0 \
(2x+1)( v'' + 4v' + 4v) - 4(x+1)( v') - 8xv - 4v &=& 0 \
end{eqnarray*}
begin{eqnarray*}
(2x+1)( v'' + 4v' ) - 4(x+1)( v') &=& 0 \
(2x+1)( v'' + 4v' ) - 4x( v') - 4 v' &=& 0 \
(2x+1)( v'') + (2x+1)(4v' ) - 4x( v') - 4 v' &=& 0 \
(2x+1)( v'') + 4xv' &=& 0 \
(2x+1)v''+ 4xv' &=& 0 \
end{eqnarray*}
Let $w = frac{dv}{dx}$.
begin{eqnarray*}
(2x+1)frac{dw}{dx} + 4xw &=& 0 \
frac{dw}{w} + frac{dx}{2x+1} &=& 0 \
int frac{dw}{w} ,, dx + int frac{dx}{2x+1} ,, dx &=& int 0 ,, dx \
end{eqnarray*}
Now to perform this integration:
$$ int frac{dx}{2x+1} ,, dx $$
we use the substitution $u_1 = 2x + 1$ which gives $du_1 = dx$.
begin{eqnarray*}
2x &=& u_1 - 1 \
4x &=& 2u_1 - 2 \
int frac{dx}{2x+1} ,, dx &=& int frac{2u_1-2}{u_1} ,, du_1 =
int 2 ,, du_1 - int frac{2}{u_1} ,, du_1 \
int frac{dx}{2x+1} ,, dx &=& 2(2x+1) - 2 ln{|2x+1|} + C_1 \
end{eqnarray*}
begin{eqnarray*}
ln{|w|} + 2(2x+1) - 2 ln{|2x+1|} + C_1 &=& 0 \
ln{|w|} + ln{e^{(4x+2)}} - 2 ln{|2x+1|} + C_1 &=& 0 \
ln{bigg|frac{we^{(4x+2)}}{(2x+1)^2}bigg|} &=& -C_1 \
frac{we^{(4x+2)}}{(2x+1)^2} &=& C_2 \
end{eqnarray*}
begin{eqnarray*}
w &=& frac{dv}{dx} = C_2 left( frac{(2x+1)^2}{e^{(4x+2)}} right) =
C_2 left( frac{(2x+1)^2}{e^{2x+1}e^{2x+1}} right) \
dv &=& frac{C_2(2x+1)^2}{e^{2x+1}e^{2x+1}} , dx \
end{eqnarray*}
Using an online integral calcuator, we find:
$$ int frac{(2x+1)^2}{e^{2x+1}e^{2x+1}} ,, dx = -frac{(8x^2+12+5)e^{-4x-2}}{8} + C_3 $$
begin{eqnarray*}
v &=& -frac{C_2(8x^2+12+5)e^{-4x-2}}{8} + C_3 \
v &=& e^{-2x}y \
e^{-2x}y &=& -frac{C_2(8x^2+12+5)e^{-4x-2}}{8} + C_3 \
end{eqnarray*}
However, the book's answer is:
$$ y = c_1e^{2x} + c_2(x+1)$$

At the point
$$(2x+1)frac{dw}{dx} + 4xw = 0
$$
you missed to transfer the factor $4x$ to the next step. It should be
$$
frac{dw}{w} + frac{4x,dx}{2x+1} = 0 ,
$$
so that then
begin{align}
frac{dw}{w} + 2,dx -frac{2,dx}{2x+1} &= 0 \
implies
ln|w|+2x-ln|2x+1|&=c,\
v'=w&=Ce^{-2x}(2x+1),\
v&=-Ce^{-2x}(x+1),\
y=e^{2x}v&=-C(x+1)
end{align}

Then, after this factor miraculously re-appears in the discussion of the $u_1$ substitution, you simply replaced $dx$ with $du_1$, while the correct formula is $u_1=2x+1implies du_1=2,dx$.

This reduction by a factor then exponent $2$ then also decreases the complexity in the ensuing computations.

(Btw., the eqnarray environment was declared obsolete decades ago. Use align and related environments. Read the official l2tabu guide in the LaTeX2e documentation.)