Welcome to our community

Be a part of something great, join today!

How to solve this Differential equation second order linear with variable coefficient

Amer

Active member
Mar 1, 2012
275
$(1-x^2)y'' - xy' + 4y =2 x \sqrt{1-x^2} $
Hint use the substitution $x =\sin t$
I used it and end with

$\cos t y'' + \sin t y' - \frac{\sin t}{\cos t} y' + 4y = 2\sin t |\cos t| $

how to solve this i just want the name of the method
 
Last edited:

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
I do not think you have gone to the $t$ domain correctly. You have
$$\frac{dy}{dt}=\frac{dy}{dx}\,\frac{dx}{dt}=\cos(t)\,\frac{dy}{dx}.$$
Then you have
$$\frac{d^{2}y}{dt^{2}}=\frac{d}{dt}\left[\cos(t)\,\frac{dy}{dx}\right]
=\cos(t)\,\frac{d}{dt}\,\frac{dy}{dx}+\frac{dy}{dx}\,(-\sin(t))$$
$$=\cos^{2}(t)\,\frac{d^{2}y}{dx^{2}}-\sin(t)\,\frac{dy}{dx}.$$
See if that doesn't do some nice things for you.
 

chisigma

Well-known member
Feb 13, 2012
1,704
$(1-x^2)y'' - xy' + 4y = x \sqrt{1-x^2} $
Hint use the substitution $x =\sin t$
I used it and end with

$\cos t y'' + \sin t y' - \frac{\sin t}{\cos t} y' + 4y = 2\sin t |\cos t| $

how to solve this i just want the name of the method
The equation is of the type complete and its solution is the sum of two terms, the general solution of the incomplete equation and any particular solution of the complete equation, so that first we have to solve the equation...

$\displaystyle (1-x^{2})\ y^{\ ''} -x\ y^{\ '} + 4\ y=0$ (1)

The procedure that I will use is a little 'non conventional' and requires a preliminary. The solution of a second order incomplete equation is of the type...


$\displaystyle y(x)= c_{1}\ u(x) + c_{2}\ v(x)$ (2)


... where $u(x)$ and $v(x)$ are two independent solutions of (1). Since u and v both satisfy (1) is...


$\displaystyle (1-x^{2})\ u^{\ ''} -x\ u^{\ '} +4\ u=0$

$\displaystyle (1-x^{2})\ v^{\ ''} -x\ v^{\ '} +4\ v=0$ (3)

Multiplying the first of (3) by v and the second by u and do the difference we obtain...

$\displaystyle (1-x^{2})\ (v\ u^{\ ''} - u\ v^{\ ''}) - x\ (v\ u^{\ '} - u\ v^{\ '}) =0$ (4)

Now we set $z= v\ u^{\ '}- u\ v^{\ '}$ so that (4) becomes...

$\displaystyle z^{\ '}= \frac {x}{1-x^{2}}\ z$ (5)

The (5) is a linear ODE the solution of which is...

$\displaystyle z=\frac{c_{2}}{\sqrt{1-x^{2}}}$ (6)

... so that is...

$\displaystyle \frac{z}{v^{2}}= \frac{d}{dx} (\frac{u}{v})= \frac{c_{2}}{v^{2}\ \sqrt{1-x^{2}}} \implies\ u= c_{1}\ v + c_{2}\ v\ \int \frac{dx}{v^{2}\ \sqrt{1-x^{2}}}$ (7)

Now it is easy enough to see that $u=2 x^{2}-1$ is solution of (1) so that from (7) we derive that...

$\displaystyle v= (2 x^{2}-1)\ \int \frac{dx}{(2 x^{2}-1)^{2}\ \sqrt{1-x^{2}}} = - x\ \sqrt{1-x^{2}}$ (8)

... is also solution of (1) so that the general solution of (1) is...

$\displaystyle y(x)= c_{1}\ (2 x^{2}-1) + c_{2}\ x\ \sqrt{1-x^{2}}$ (9)

... and half of the work is done. The second half will be done [if possible...] in next post...

Kind regards

$\chi$ $\sigma$
 
Last edited:

Amer

Active member
Mar 1, 2012
275
I do not think you have gone to the $t$ domain correctly. You have
$$\frac{dy}{dt}=\frac{dy}{dx}\,\frac{dx}{dt}=\cos(t)\,\frac{dy}{dx}.$$
Then you have
$$\frac{d^{2}y}{dt^{2}}=\frac{d}{dt}\left[\cos(t)\,\frac{dy}{dx}\right]
=\cos(t)\,\frac{d}{dt}\,\frac{dy}{dx}+\frac{dy}{dx}\,(-\sin(t))$$
$$=\cos^{2}(t)\,\frac{d^{2}y}{dx^{2}}-\sin(t)\,\frac{dy}{dx}.$$
See if that doesn't do some nice things for you.
What i did My question is

[tex](1-x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} + 4y =2 x \sqrt{1-x^2} [/tex]

The sub is
[tex] x = \sin t \Rightarrow \dfrac{dx}{dt} = \cos t \Rightarrow \dfrac{d^2x}{dt^2} = -\sin t [/tex]

[tex]\dfrac{dy}{dt} = \dfrac{dy}{dx} \dfrac{dx}{dt} [/tex]

I made a mistake in follwing, i write it like below but when i differentiate with respect to t,
[tex]\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} [/tex]
the left hand side I write it like this
[tex]\dfrac{d^2y}{dx^2} [/tex] which should be like this [tex]\dfrac{d^2y}{dx^2} \dfrac{dy}{dt} [/tex]

Thanks very much both
 

Amer

Active member
Mar 1, 2012
275
toi
The equation is of the type complete and its solution is the sum of two terms, the general solution of the incomplete equation and any particular solution of the complete equation, so that first we have to solve the equation...

$\displaystyle (1-x^{2})\ y^{\ ''} -x\ y^{\ '} + 4\ y=0$ (1)

The procedure that I will use is a little 'non conventional' and requires a preliminary. The solution of a second order incomplete equation is of the type...


$\displaystyle y(x)= c_{1}\ u(x) + c_{2}\ v(x)$ (2)


... where $u(x)$ and $v(x)$ are two independent solutions of (1). Since u and v both satisfy (1) is...


$\displaystyle (1-x^{2})\ u^{\ ''} -x\ u^{\ '} +4\ u=0$

$\displaystyle (1-x^{2})\ v^{\ ''} -x\ v^{\ '} +4\ v=0$ (3)

Multiplying the first of (3) by v and the second by u and do the difference we obtain...

$\displaystyle (1-x^{2})\ (v\ u^{\ ''} - u\ v^{\ ''}) - x\ (v\ u^{\ '} - u\ v^{\ '}) =0$ (4)

Now we set $z= v\ u^{\ '}- u\ v^{\ '}$ so that (4) becomes...

$\displaystyle z^{\ '}= \frac {x}{1-x^{2}}\ z$ (5)

The (5) is a linear ODE the solution of which is...

$\displaystyle z=\frac{c_{2}}{\sqrt{1-x^{2}}}$ (6)

... so that is...

$\displaystyle \frac{z}{v^{2}}= \frac{d}{dx} (\frac{u}{v})= \frac{c_{2}}{v^{2}\ \sqrt{1-x^{2}}} \implies\ u= c_{1}\ v + c_{2}\ v\ \int \frac{dx}{v^{2}\ \sqrt{1-x^{2}}}$ (7)

Now it is easy enough to see that $u=x$ is solution of (1) so that from (7) we derive that...

$\displaystyle v= x\ \int \frac{dx}{x^{2}\ \sqrt{1-x^{2}}} = - \sqrt{1-x^{2}}$ (7)

... is also solution of (1) so that the general solution of (1) is...

$\displaystyle y(x)= c_{1}\ x + c_{2}\ \sqrt{1-x^{2}}$ (8)

... and half of the work is done. The second half will be done [if possible...] in next post...

Kind regards

$\chi$ $\sigma$
I am a bit new to differential equation I do not know what complete and incomplete equations mean
can you give me a link about it or a little explanation ?
Thanks
 

chisigma

Well-known member
Feb 13, 2012
1,704
toi

I am a bit new to differential equation I do not know what complete and incomplete equations mean
can you give me a link about it or a little explanation ?
Thanks
An incomplete second order linear ODE is written as...

$\displaystyle a(x)\ y^{\ ''} + b(x)\ y^{\ '} + c(x)\ y=0$ (1)

... and a complete second order linear ODE as...


$\displaystyle a(x)\ y^{\ ''} + b(x)\ y^{\ '} + c(x)\ y= d(x)$ (2)

If u(x) and v(x) are independent solutions of (1) and w(x) is any particular solution of (2), the the general solution of (2) is...

$\displaystyle y(x)= c_{1}\ u(x) + c_{2}\ v(x) + w(x)$ (3)

... where $c_{1}$ and $c_{2}$ are arbitrary constants...

Kind regards

$\chi$ $\sigma$
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
If you're clever enough (or lucky enough) to guess one solution of the homogeneous problem

$(1-x^2)y'' - xy' + 4y = 0$ in this case $y = 2x^2-1$

then $y = (2x^2-1)v$ will reduce your ODE to one that is second order with the $v$ term missing and letting $w = v'$ will then give you one that is linear in $w$!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
An incomplete second order linear ODE is written as...

$\displaystyle a(x)\ y^{\ ''} + b(x)\ y^{\ '} + c(x)\ y=0$ (1)

... and a complete second order linear ODE as...


$\displaystyle a(x)\ y^{\ ''} + b(x)\ y^{\ '} + c(x)\ y= d(x)$ (2)

If u(x) and v(x) are independent solutions of (1) and w(x) is any particular solution of (2), the the general solution of (2) is...

$\displaystyle y(x)= c_{1}\ u(x) + c_{2}\ v(x) + w(x)$ (3)

... where $c_{1}$ and $c_{2}$ are arbitrary constants...

Kind regards

$\chi$ $\sigma$
For translation effectiveness: chi sigma's "incomplete" is often termed "homogeneous", and chi sigma's "complete" is often termed "inhomogeneous".
 

chisigma

Well-known member
Feb 13, 2012
1,704
For translation effectiveness: chi sigma's "incomplete" is often termed "homogeneous", and chi sigma's "complete" is often termed "inhomogeneous".
The reason why I prefer the terms 'complete' and 'incomplete' is that these terms are related to the presence or not of the 'known term' d(x). It is obvious that the correspondence 'homogeneous -> incomplete' and 'inhomogeneous -> complete' is a perfect source of confusion...


Kind regards


$\chi$ $\sigma$
 

chisigma

Well-known member
Feb 13, 2012
1,704
My attempts to do the second part of the work searching a particular solution of the complete equation...

$\displaystyle (1-x^{2})\ y^{\ ''} -x\ y^{\ '} + 4\ y= 2\ x\ \sqrt{1-x^{2}}$ (1)

... didn't produce results but the problem would be easily overcome changing in (1) the sigh of the term in y' so that the ODE becomes...

$\displaystyle (1-x^{2})\ y^{\ ''} + x\ y^{\ '} + 4\ y= 2\ x\ \sqrt{1-x^{2}}$ (2)

I would ask Amer if May be that the equation is (2) and not (1)...


Kind regards


$\chi$ $\sigma$
 

Amer

Active member
Mar 1, 2012
275
My attempts to do the second part of the work searching a particular solution of the complete equation...

$\displaystyle (1-x^{2})\ y^{\ ''} -x\ y^{\ '} + 4\ y= 2\ x\ \sqrt{1-x^{2}}$ (1)

... didn't produce results but the problem would be easily overcome changing in (1) the sigh of the term in y' so that the ODE becomes...

$\displaystyle (1-x^{2})\ y^{\ ''} + x\ y^{\ '} + 4\ y= 2\ x\ \sqrt{1-x^{2}}$ (2)

I would ask Amer if May be that the equation is (2) and not (1)...


Kind regards


$\chi$ $\sigma$
no i write the question correctly it is -xy'
 

chisigma

Well-known member
Feb 13, 2012
1,704
The ODE originally posted by Amer [slighty modified by me...] was...

$\displaystyle y'' - \frac{x}{1-x^{2}}\ y' + \frac{4}{1-x^{2}}\ y =\frac{2\ x}{\sqrt{1-x^2}}$ (1)

... and in the post #3 we found the general solution of the incomplete equation…

$\displaystyle y'' - \frac{x}{1-x^{2}}\ y' + \frac{4}{1-x^{2}}\ y = 0$ (2)

... that is...

$\displaystyle y(x)= c_{1}\ (2\ x^{2} -1) + c_{2}\ x\ \sqrt{1-x^{2}}$ (3)

Now we have to search a particular solution of the (1) and that will be performed describing a general procedure for finding a particular solution of an ODE like...

$\displaystyle y^{\ ''} + p(x)\ y^{\ '} + q(x)\ y = \varphi (x)$ (4)

Let’s suppose to know the general solution of the incomplete equation…

$\displaystyle y^{\ ''} + p(x)\ y^{\ '} + q(x)\ y = 0$ (5)

… that is…

$\displaystyle y(x)= c_{1}\ u(x) + c_{2}\ v(x)$ (6)

... and to write the particular solution of the complete equation like...

$\displaystyle Y(x)= C_{1}(x)\ u(x) + C_{2} (x)\ v(x)$ (7)

... where $C_{1} (x)$ and $C_{2} (x)$ are functions of x. In...

Second Order Linear Nonhomogeneous Differential Equations with Variable Coefficients

... is demonstrated that $C_{1} (x)$ and $C_{2} (x)$ have the form...


$\displaystyle C_{1} (x)= - \int \frac{ v (x)\ \varphi(x)} {W_{u,v} (x)}\ dx$

$\displaystyle C_{2} (x)= \int \frac{ u (x)\ \varphi(x)} {W_{u,v} (x)}\ dx$ (8)

... where $\displaystyle W_{u,v} (x) = u(x)\ v^{\ '} (x) - v(x)\ u^{\ '}(x)$ is the Wronskian of u and v. Now we turn back to (1), remembering that in previous post we found...

$\displaystyle u = 2\ x^{2}-1 \implies u^{\ '}= 4\ x$

$\displaystyle v = x\ \sqrt{1-x^{2}} \implies v^{\ '} = \sqrt{1-x^{2}} - \frac{x^{2}}{\sqrt{1-x^{2}}}$ (9)

... so that is ...

$\displaystyle W_{u,v} (x)= - (2\ x^{2} +1)\ \sqrt{1-x^{2}} - \frac{ x^{2}\ (2\ x^{2}-1)} {\sqrt{1-x^{2}}}= - \frac{1}{\sqrt{1-x^{2}}}$ (10)

... and remembering that is $\displaystyle \varphi(x)= \frac{2\ x}{\sqrt{1-x^{2}}}$ we finally obtain...

$\displaystyle C_{1}(x)= 2\ \int x^{2}\ \sqrt{1-x^{2}}\ dx = \frac{1}{4}\ \{ x\ \sqrt{1-x^{2}}\ (2\ x^{2}-1) + \sin^{-1} x\}$

$\displaystyle C_{2}(x)= 2\ \int (x - 2\ x^{2})\ dx = x^{2}\ (1-x^{2})$ (11)

... and (7) becomes...

$\displaystyle Y(x)= \frac{2\ x^{2}-1}{4}\ \{x\ \sqrt{1-x^{2}}\ (2\ x^{2}-1) + \sin^{-1} x\} + ( x\ \sqrt{1-x^{2}})^{3}$ (12)

Of course the 'old wolf' isn't a 'Superman' in pure calculus so that it is better that some 'young mind' controls these results (Wasntme)...

Kind regards


$\chi$ $\sigma$
 
Last edited:

JJacquelin

New member
Jul 31, 2012
3
Hi !

I suggest to bring back your result y(x) into the ODE. So that, you will check if your result is correct or not.
This is what I obtained :
 

Attachments

chisigma

Well-known member
Feb 13, 2012
1,704
Welcome on MHB JJaquelin!... after Your post I verified some mistakes in my calculation and I corrected them... anyway that is probably not jet all right and further controls of me are necessary...


Kind regards


$\chi$ $\sigma$