Question

Problem:
y = tan(2arctanx)

prove that:

(1+x^2)y'' = 2(2y-x)y'

I manged to get another solution for it, but I wanna know what's wrong with this solution.
or how to complete it.

y = 2tan(arctanx)/1-tan^2(arctanx)

= 2x/1 - x^2

(1-x^2)y = 2x

-2xy + (1-x^2)y' = 2

-2y -2xy' + (-2x)y' + (1-x^2)y'' = 0

(1-x^2)y'' = 2(y - 2x)y'

Answer 1

here we have to develop explicitly the indicated computations

Answer 2

Correct, after deriving the given identity , I got to the proof easily. However , The other solution looked easier to me, when I thought of the problem. as I won't have to derive any inverse or trigonometric function.

Answer 3

Do you think that there is a way to turn

(1-x^2)y'' = 2(y - 2x)y'

into:

(1+x^2)y'' = 2(2y-x)y'

Answer 4

at first sight I don't know, I think that the best way in order to solve your problem is a direct substitution of the function \(y\) into the ODE

Answer 5

See this too:

y = tan^'1 cotx^2 + sin^2cos-1sqrt(x)

Using cos^-1 triangle I got that

sin^-1 = sqrt(1-x)

y = 1/x^2 + 1-x

y' = -2x^-3 -1

while I must prove that

y' = -2x - 1

I seems like, trying to simplify it, complicates it instead.

Answer 6

for example, if we use the chain rule, we can write this:
\[\frac{{dy}}{{dx}} = \frac{{dy}}{{dz}} \cdot \frac{{dz}}{{dx}} = \frac{d}{{dz}}\left( {\tan z} \right) \cdot \frac{d}{{dx}}\left( {2\arctan x} \right)\]
as you can see, I used this new variable: \(z=2 \arctan x\)

Answer 7

and I wrote:
\(y=\tan z\)

Answer 8

so, after a simple computation, we get:
\[\begin{gathered}
\frac{{dy}}{{dx}} = \frac{{dy}}{{dz}} \cdot \frac{{dz}}{{dx}} = \frac{d}{{dz}}\left( {\tan z} \right) \cdot \frac{d}{{dx}}\left( {2\arctan x} \right) = \hfill \\
\hfill \\
= \left( {{{\left( {\tan z} \right)}^2} + 1} \right) \cdot \frac{2}{{1 + {x^2}}} = \hfill \\
\hfill \\
= \left( {{{\left( {\tan \left( {2\arctan x} \right)} \right)}^2} + 1} \right) \cdot \frac{2}{{1 + {x^2}}} \hfill \\
\end{gathered} \]

Answer 9

y' = sec^2(2arctan(x)) * 2 * 1/(1+x^2)

y' = (1+ tan^2(2arctan(x)) * 2/(1+x^2)

(1+x^2)y' = 2(1+y^2)

(1+x^2)y'' + 2xy' = 4yy'

(1+x^2)y'' = 2(2y - x)y'

Answer 10

The problem is, That confuses me,each time, I have to re-solve the problem twice:

1) with simplification
2) without

Answer 11

I don't understand, now we have computed the first derivative, and then we have differentiated the so obtained ODE, so we got the requested ODE, it is a correct procedure

Answer 12

See this y = tan(2arctanx),

Why would I differentiate tan, tan inverse when I could differentiate 2x/1+x^2 ?

Answer 13

another different procedure is to compute the second derivative, so substituting both the first and the second derivatives into the given ODE, we have to obtain an identity

Answer 14

since we can consider that function as a function of function, so we can apply the so called \(chain\;rule\)
as I wrote before, we can write this:
\(y=\tan z\) and \(z=2 \arctan x\), so if we consider the composed function, we get:
\(y= \tan z= \tan ( 2 \arctan x)\)
and the chain rule is:
\[\frac{{dy}}{{dx}} = \frac{{dy}}{{dz}} \cdot \frac{{dz}}{{dx}}\]

Answer 15

I do chain rule all in one.

y' = sec^2(2arctan(x)) * 2 * 1/(1+x^2)

y' = (1+ tan^2(2arctan(x)) * 2/(1+x^2)

(1+x^2)y' = 2(1+y^2)

(1+x^2)y'' + 2xy' = 4yy'

(1+x^2)y'' = 2(2y - x)y'

Answer 16

correct! your first step is just the \(chain\; rule\)

Answer 17

Why when I simplify it , I don't get what I want -,-

Answer 18

I see that all your last steps are correct!

Answer 19

y = 2tan(arctanx)/1-tan^2(arctanx)

= 2x/1 - x^2

(1-x^2)y = 2x

-2xy + (1-x^2)y' = 2

-2y -2xy' + (-2x)y' + (1-x^2)y'' = 0

(1-x^2)y'' = 2(y - 2x)y'

Answer 20

I'm checking your steps, please wait...

Answer 21

Take your time.

Answer 22

I'm not able to get the right result, I don't know, I continue to try, and tomorrow I will write my reasonings.

Answer 23

oops.. I don't succeed to get the right result...

Answer 24

Thanks so much for your time.

Answer 25

:)

Answer 26

If you succeeded , post the result, here and tag me.

Answer 27

ok! :)

Answer 28

here is my resoning:
using the function:
\[y = \frac{{2x}}{{1 - {x^2}}}\]
I computed both first and second derivatives:
\[y' = 2\frac{{1 + {x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}},\quad y'' = \frac{{4x\left( {3 + {x^2}} \right)}}{{{{\left( {1 - {x^2}} \right)}^3}}}\]
using such derivative, we can show that left side = right side of your ODE, and both sides are equal to:
\[\frac{{4x\left( {3 + {x^2}} \right)\left( {1 + {x^2}} \right)}}{{{{\left( {1 - {x^2}} \right)}^3}}}\]

Answer 29

@TrojanPoem

Answer 30

Nice job, @Michele_Laino .

Answer 31

thanks!! :) @TrojanPoem