Finding the solution to a second order ordinary differential equation where one solution is already known

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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:
$$ y = c_1e^{2x} + c_2(x+1)$$










share|cite|improve this question











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    $begingroup$


    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:
    $$ y = c_1e^{2x} + c_2(x+1)$$










    share|cite|improve this question











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      1












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      1





      $begingroup$


      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:
      $$ y = c_1e^{2x} + c_2(x+1)$$










      share|cite|improve this question











      $endgroup$




      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:
      $$ y = c_1e^{2x} + c_2(x+1)$$







      ordinary-differential-equations






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      edited Dec 10 '18 at 12:14







      Bob

















      asked Dec 9 '18 at 23:09









      BobBob

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          $begingroup$

          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.)






          share|cite|improve this answer











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            $begingroup$

            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.)






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              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.)






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                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.)






                share|cite|improve this answer











                $endgroup$



                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.)







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 9 '18 at 23:43

























                answered Dec 9 '18 at 23:18









                LutzLLutzL

                59.8k42057




                59.8k42057






























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