Question

Find a second solution y2 for (2x+1)xy"-2(2x^2-1)y'-4(x+1)y=0; y1=1/x that isn't a constant multiple of the solution y1.

Answer 1

\[y _{2}=\frac{ u }{ x }\]

Answer 2

\[y _{2}'=\frac{ u' }{ x }-\frac{ u }{ x^2 }\]

Answer 3

\[y _{2}''=\frac{ u'' }{ x }-\frac{ 2u' }{ x^2 }+\frac{ 2u }{ x^3 }\]

Answer 4

@SithsAndGiggles

Answer 5

i think its 0

Answer 6

nvm

Answer 7

Plug in the derivatives:
\[\begin{align*}
0&=(2x+1)xy''-2(2x^2-1)y'-4(x+1)y\\\\
&=(2x+1)x\left(\frac{u''}{x}-\frac{2u'}{x^2}+\frac{2u}{x^3}\right)-2(2x^2-1)\left(\frac{u'}{x}-\frac{u}{x^2}\right)-4(x+1)\frac{u}{x}\\\\
&=(2x+1)u''-\frac{2(2x+1)}{x}u'+\frac{2(2x+1)}{x^2}u-\frac{2(2x^2-1)}{x}u'\\
&\quad\quad+\frac{2(2x^2-1)}{x^2}u-4\frac{x+1}{x}u\\\\
&=(2x+1)u''-(4x+4)u'
\end{align*}\]
Substituting \(t=u'\) gives a linear ODE,
\[(2x+1)t'-4(x+1)t=0\]

Answer 8

Thank you!

Answer 9

But at the end, I got \[y _{2}=\frac{ u }{ x }=\frac{ C _{1}e ^{x/2}(4x-6)+C _{2} }{ x }\]

Answer 10

The answer in the book is \[y _{2}=e ^{2x}\].

Answer 11

Hmm, checking with Mathematica says \(e^{2x}\) should be the solution, but our work thus far gives \((2x+1)e^{2x}\) is... The algebra checks out, so I'm not sure what the problem is.

Answer 12

So how do we get the right solution e^(2x)?

Answer 13

I'll look into it and report back when I find out :)

Answer 14

Okay.

Answer 15

It doesn't look like any of my work is wrong, so I'm thinking you made a mistake somewhere when you solved for \(u\).

Let's return to the linear ODE in terms of \(t=u'\):
\[(2x+1)t-4(x+1)t=0~~\iff~~t'-4\frac{x+1}{2x+1}t=0\]
Separating the variables gives
\[\frac{dt}{t}=4\frac{x+1}{2x+1}\,dx\]
and integrating gives
\[\ln|t|=(2x+1)+\ln|2x+1|+C_1~~\implies~~t=C_1(2x+1)e^{2x+1}\]
Integrating again (to solve for \(u\)) gives
\[u=\int C_1(2x+1)e^{2x+1}\,dx=\frac{C_1}{2}(2x+1)e^{2x+1}-\frac{C_1}{2}e^{2x+1}+C_2\]
which reduces to
\[u=C_1xe^{2x+1}+C_2=C_1xe^{2x}+C_2\]
Dividing by \(x\) to solve for \(y_2\) gives what we want:
\[y=\frac{u}{x}~~\implies~~y_2=e^{2x}\]

Answer 16

Thank you! I finally got this problem!