Welcome to our community

Be a part of something great, join today!

second order differential equation,with constant terms

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hello (Smirk)
Given the [tex] x^{2}y''+axy'+by=0[/tex],I have to show that with replacing [tex] x[/tex] with [tex]e^{z}[/tex],it becomes a second order differential equation,with constant terms.
I tried to do this and I got this: [tex] y''+\frac{a}{e^{z}}y'+\frac{b}{e^{2z}}y=0 [/tex].
But,at this equation the terms aren't constant :confused: What else could I do??
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
Hello (Smirk)
Given the [tex] x^{2}y''+axy'+by=0[/tex],I have to show that with replacing [tex] x[/tex] with [tex]e^{z}[/tex],it becomes a second order differential equation,with constant terms.
I tried to do this and I got this: [tex] y''+\frac{a}{e^{z}}y'+\frac{b}{e^{2z}}y=0 [/tex].
But,at this equation the terms aren't constant :confused: What else could I do??
Hi evinda!

Note that $y$ actually means $y(x)$, and $y'$ actually means \(\displaystyle \frac{dy}{dx}\).
You did not replaces those $x$'s yet.


Suppose we define $Y(z) = y(x(z)) = y(e^z)$.
Then according to the chain rule:
$$Y'(z) = \frac{dY(z)}{dz} = \frac{dy(x(z))}{dz} = \frac{dy(x)}{dx} \frac{dx(z)}{dz} = y'(x) \frac{dx(z)}{dz}$$
Or shorter:
$$Y' = \frac{dY}{dz} = \frac{dy}{dx} \frac{dx}{dz} = y' \frac{dx}{dz}$$
Perhaps you can express your differential equation with Y, Y', and Y''?


Btw, I have moved your thread to the sub forum Differential Equations, since that is the topic at hand.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hi evinda!

Note that $y$ actually means $y(x)$, and $y'$ actually means \(\displaystyle \frac{dy}{dx}\).
You did not replaces those $x$'s yet.


Suppose we define $Y(z) = y(x(z)) = y(e^z)$.
Then according to the chain rule:
$$Y'(z) = \frac{dY(z)}{dz} = \frac{dy(x(z))}{dz} = \frac{dy(x)}{dx} \frac{dx(z)}{dz} = y'(x) \frac{dx(z)}{dz}$$
Or shorter:
$$Y' = \frac{dY}{dz} = \frac{dy}{dx} \frac{dx}{dz} = y' \frac{dx}{dz}$$
Perhaps you can express your differential equation with Y, Y', and Y''?


Btw, I have moved your thread to the sub forum Differential Equations, since that is the topic at hand.
I found this:
[tex] x^{2}y''\frac{dx}{dz}+x^{2}y'\frac{d^{2}x}{dz^{2}}+axy'\frac{dx}{dz}+by=0 [/tex]
Is this right??If yes,how can I continue?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
I found this:
[tex] x^{2}y''\frac{dx}{dz}+x^{2}y'\frac{d^{2}x}{dz^{2}}+axy'\frac{dx}{dz}+by=0 [/tex]
Is this right??If yes,how can I continue?
Looks like you substituted it the wrong way around.

From:
$$Y' = y' \frac{dx}{dz}$$
we get:
$$y' = \frac{Y'}{\frac{dx}{dz}} = \frac{Y'}{e^z} = \frac{Y'}{x}$$
Perhaps you can substitute that in the original DE?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Looks like you substituted it the wrong way around.

From:
$$Y' = y' \frac{dx}{dz}$$
we get:
$$y' = \frac{Y'}{\frac{dx}{dz}} = \frac{Y'}{e^z} = \frac{Y'}{x}$$
Perhaps you can substitute that in the original DE?
So,is it like that:

[tex] y''=\frac{y''(z)}{\frac{dx}{dz}}-y'(z) [/tex] ?
 

chisigma

Well-known member
Feb 13, 2012
1,704
Hello (Smirk)
Given the [tex] x^{2}y''+axy'+by=0[/tex],I have to show that with replacing [tex] x[/tex] with [tex]e^{z}[/tex],it becomes a second order differential equation,with constant terms.
I tried to do this and I got this: [tex] y''+\frac{a}{e^{z}}y'+\frac{b}{e^{2z}}y=0 [/tex].
But,at this equation the terms aren't constant :confused: What else could I do??
The standard way to transform an ODE of this type [Euler-Cauchy differential equation...] is in the substitution $u = \ln x$, so that You have...


$\displaystyle \frac{d y}{d x} = \frac{d y}{d u} \frac{d u} {d x} = \frac{1}{x} \ \frac{dy}{d u}\ (1)$

$\displaystyle \frac{d^{2} y}{d x^{2}} = \frac{d^{2} y}{d u^{2}} (\frac{d u}{d x})^{2} + \frac{d y}{d u} \frac{d^{2} u}{d x^{2}} = \frac{1}{x^{2}} (\frac{d^{2} y}{d u^{2}} - \frac{d y}{d u})\ (2)$


Inserting (1) and (2) into the original ODE You obtain...

$\displaystyle \frac{d^{2} y}{d u^{2}} + (a -1)\ \frac{d y}{d u} + b\ y =0\ (3)$

A 'very pratical' way to attack this type of equation is however to search solutions of the form $y = x^{\nu}$. Imposing that You arrive at a second order algebraic equation in $\nu$ and if $\nu_{1}$ and $\nu_{2}$ are the solutions, then the general solution of the ODE is...

$\displaystyle y(x) = c_{1}\ x^{\nu_{1}} + c_{2}\ x^{\nu_{2}}\ (4)$

Kind regards

$\chi$ $\sigma$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
The standard way to transform an ODE of this type [Euler-Cauchy differential equation...] is in the substitution $u = \ln x$, so that You have...


$\displaystyle \frac{d y}{d x} = \frac{d y}{d u} \frac{d u} {d x} = \frac{1}{x} \ \frac{dy}{d u}\ (1)$

$\displaystyle \frac{d^{2} y}{d x^{2}} = \frac{d^{2} y}{d u^{2}} (\frac{d u}{d x})^{2} + \frac{d y}{d u} \frac{d^{2} u}{d x^{2}} = \frac{1}{x^{2}} (\frac{d^{2} y}{d u^{2}} - \frac{d y}{d u})\ (2)$


Inserting (1) and (2) into the original ODE You obtain...

$\displaystyle \frac{d^{2} y}{d u^{2}} + (a -1)\ \frac{d y}{d u} + b\ y =0\ (3)$

A 'very pratical' way to attack this type of equation is however to search solutions of the form $y = x^{\nu}$. Imposing that You arrive at a second order algebraic equation in $\nu$ and if $\nu_{1}$ and $\nu_{2}$ are the solutions, then the general solution of the ODE is...

$\displaystyle y(x) = c_{1}\ x^{\nu_{1}} + c_{2}\ x^{\nu_{2}}\ (4)$

Kind regards

$\chi$ $\sigma$
I understand :) But how from this equation: $\displaystyle \frac{d^{2} y}{d u^{2}} + (a -1)\ \frac{d y}{d u} + b\ y =0\ $ can I get the general solution??
I tried like that: [tex] r^{2}+(a-1)r+b=0 =>
d=(a-1)^{2}-4b
[/tex]
but I don't know how to continue...
 

chisigma

Well-known member
Feb 13, 2012
1,704
I understand :) But how from this equation: $\displaystyle \frac{d^{2} y}{d u^{2}} + (a -1)\ \frac{d y}{d u} + b\ y =0\ $ can I get the general solution??
I tried like that: [tex] r^{2}+(a-1)r+b=0 =>
d=(a-1)^{2}-4b
[/tex]
but I don't know how to continue...
The only You have to do is to complete the solution of the second order equation...

$\displaystyle r^{2}+(a-1)r + b = 0 \implies r_{1} = \frac{1 - a - \sqrt{(1-a)^{2} - 4 b}}{2},\ r_{2} = \frac{1 - a + \sqrt{(1-a)^{2} - 4 b}}{2}\ (1)$

... and the solution is given by...

$\displaystyle y = c_{1} e^{r_{1}\ u} + c_{2} e^{r_{2}\ u} = c_{1}\ x^{r_{1}} + c_{2}\ x^{r_{2}}\ (2)$

Kind regards

$\chi$ $\sigma$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
The only You have to do is to complete the solution of the second order equation...

$\displaystyle r^{2}+(a-1)r + b = 0 \implies r_{1} = \frac{1 - a - \sqrt{(1-a)^{2} - 4 b}}{2},\ r_{2} = \frac{1 - a + \sqrt{(1-a)^{2} - 4 b}}{2}\ (1)$

... and the solution is given by...

$\displaystyle y = c_{1} e^{r_{1}\ u} + c_{2} e^{r_{2}\ u} = c_{1}\ x^{r_{1}} + c_{2}\ x^{r_{2}}\ (2)$

Kind regards

$\chi$ $\sigma$
I got it!!!And if I want to look at the same problem for x<0,what do I have to do???Maybe to set [tex] x=-e^{u} [/tex],or am I wrong?
 

chisigma

Well-known member
Feb 13, 2012
1,704
I got it!!!And if I want to look at the same problem for x<0,what do I have to do???Maybe to set [tex] x=-e^{u} [/tex],or am I wrong?
That is a very interesting question!... we have seen that the solution is of the type...

$\displaystyle y(x) = c_{1}\ x^{r_{1}} + c_{2}\ x^{r_{2}}\ (1)$

A function like $x^{r}$, where r may be any real [or even complex...] number, in general has in x=0 a singularity called brantch point, and that means that from x=0 several brantches of the function merge. For x<0 the function has several brantches and in general has a real and an imaginary part. An interesting example is the function $x^{\sqrt{2}}$ plotted by 'MonsterWolfram'...

x^(sqrt(2)) from -1 to 1 - Wolfram|Alpha

Kind regards

$\chi$ $\sigma$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
That is a very interesting question!... we have seen that the solution is of the type...

$\displaystyle y(x) = c_{1}\ x^{r_{1}} + c_{2}\ x^{r_{2}}\ (1)$

A function like $x^{r}$, where r may be any real [or even complex...] number, in general has in x=0 a singularity called brantch point, and that means that from x=0 several brantches of the function merge. For x<0 the function has several brantches and in general has a real and an imaginary part. An interesting example is the function $x^{\sqrt{2}}$ plotted by 'MonsterWolfram'...

x^(sqrt(2)) from -1 to 1 - Wolfram|Alpha

Kind regards

$\chi$ $\sigma$
So,what do I have to do to show that [tex] x^{2}y''+axy'+by=0[/tex] with [tex]x<0 [/tex] becomes a second order differential equation,with constant terms? :confused:
 

chisigma

Well-known member
Feb 13, 2012
1,704
So,what do I have to do to show that [tex] x^{2}y''+axy'+by=0[/tex] with [tex]x<0 [/tex] becomes a second order differential equation,with constant terms? :confused:
The solving procedure is valid for any value of $- \infty < x < + \infty$... for x<0 there are only some [minor] problems...

Kind regards

$\chi$ $\sigma$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
The solving procedure is valid for any value of $- \infty < x < + \infty$... for x<0 there are only some [minor] problems...

Kind regards

$\chi$ $\sigma$
Don't I have to set x to something negative?
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720