MHB Second order differential equation,with constant terms

[evinda]

Hello (Smirk)
Given the x^{2}y''+axy'+by=0,I have to show that with replacing x with e^{z},it becomes a second order differential equation,with constant terms.
I tried to do this and I got this: y''+\frac{a}{e^{z}}y'+\frac{b}{e^{2z}}y=0.
But,at this equation the terms aren't constant What else could I do??

 

[I like Serena]

Hello (Smirk)
Given the x^{2}y''+axy'+by=0,I have to show that with replacing x with e^{z},it becomes a second order differential equation,with constant terms.
I tried to do this and I got this: y''+\frac{a}{e^{z}}y'+\frac{b}{e^{2z}}y=0.
But,at this equation the terms aren't constant What else could I do??

Hi evinda!

Note that $y$ actually means $y(x)$, and $y'$ actually means $$\frac{dy}{dx}$$.
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]

Hi evinda!

Note that $y$ actually means $y(x)$, and $y'$ actually means $$\frac{dy}{dx}$$.
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:
x^{2}y''\frac{dx}{dz}+x^{2}y'\frac{d^{2}x}{dz^{2}}+axy'\frac{dx}{dz}+by=0
Is this right??If yes,how can I continue?

 

[I like Serena]

I found this:
x^{2}y''\frac{dx}{dz}+x^{2}y'\frac{d^{2}x}{dz^{2}}+axy'\frac{dx}{dz}+by=0
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]

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:

y''=\frac{y''(z)}{\frac{dx}{dz}}-y'(z) ?

 

[MarkFL]

I would suggest taking a look at this thread:

http://mathhelpboards.com/questions-other-sites-52/holmes-question-yahoo-answers-regarding-cauchy-euler-equation-3241.html

 

[chisigma]

Hello (Smirk)
Given the x^{2}y''+axy'+by=0,I have to show that with replacing x with e^{z},it becomes a second order differential equation,with constant terms.
I tried to do this and I got this: y''+\frac{a}{e^{z}}y'+\frac{b}{e^{2z}}y=0.
But,at this equation the terms aren't constant 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]

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: r^{2}+(a-1)r+b=0 =&gt;<br /> d=(a-1)^{2}-4b<br />
but I don't know how to continue...

 

[chisigma]

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: r^{2}+(a-1)r+b=0 =&gt;<br /> d=(a-1)^{2}-4b<br />
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]

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 x=-e^{u},or am I wrong?

 

[chisigma]

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 x=-e^{u},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]

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 x^{2}y&#039;&#039;+axy&#039;+by=0 with x&lt;0 becomes a second order differential equation,with constant terms?

 

[chisigma]

So,what do I have to do to show that x^{2}y&#039;&#039;+axy&#039;+by=0 with x&lt;0 becomes a second order differential equation,with constant terms?

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]

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]

Don't I have to set x to something negative?

Or can I just set x=-p,p>0??