You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
DE need help with the substitution.
- Thread starter Sonal7
- Start date
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
- Messages
- 7,763
What you wrote is difficult to read- particularly since you have "\(\displaystyle \frac{dy}{du}\)" without having defined u!
The equation is \(\displaystyle x^2\frac{d^2y}{dx^2}+ 6x\frac{dy}{dx}+ 4y= 0\)
That is what is known as an "Euler-type" (because everything is named for Euler!) or "equi-potential" (because the power of x in each term is equal to the order of the derivative). There are typically taught in an introductory Differential Equations course soon after "equations with constant coefficients" because the substitution u= ln(x) (That's what you meant to say, right) converts it to an equation with constant coefficients. That is, I think, what you are trying to do.
With \(\displaystyle u= ln(x)\), \(\displaystyle \frac{du}{dx}= \frac{1}{x}\) so \(\displaystyle \frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}= \frac{1}{x}\frac{dy}{du}\).
And then \(\displaystyle \frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{du}\right)= \frac{1}{x}\frac{d}{dx}\left(\frac{dy}{du}\right)- \frac{1}{x^2}\frac{dy}{du}= \frac{1}{x^2}\frac{d^2y}{du^2}- \frac{1}{x^2}\frac{dy}{du}\).
Putting those into the equation we have \(\displaystyle \frac{d^2y}{du^2}- \frac{dy}{du}+ 6\frac{dy}{du}+ 4y= \frac{d^2y}{du^2}+ 5\frac{dy}{du}+ 4y= 0\).
That equation has "characteristic equation" \(\displaystyle r^2+ 5r+ 4= (r+ 4)(r+ 1)= 0\) so characteristic roots r= -4 and r= -1. The general solution to this equation is \(\displaystyle y(u)= Ae^{-4u}+ Be^{-u}\). Since u= ln(x), \(\displaystyle y(x)= Ae^{-4ln(x)}+ Be^{-ln(x)}= Ae^{ln(x^{-4}}+ Be^{ln(x^{-1}}= Ax^{-4}+ Bx^{-1}= \frac{A}{x^4}+ \frac{B}{x}\).
(Some texts introduce these equations with the suggestion that you try solutions of the form \(\displaystyle y= x^r\). Had we done that here, since \(\displaystyle y'= rx^{r-1}\) and \(\displaystyle y''= r(r-1)x^{r-2}\) the equation becomes \(\displaystyle r(r-1)x^r+ 6rx^r+ 4x^r= (r^2+ 5r+ 4)x^r= 0\). For this to be true for all x, we must have, as before, \(\displaystyle r^2+ 5r+ 1= (r+ 4)(r+ 1)= 0\) so the two independent solutions are \(\displaystyle x^-4\) and \(\displaystyle x^{-1}\) giving the same solutions. The difficulty is that multiple roots or complex roots of the characteristic equation make this more complicated!)
The equation is \(\displaystyle x^2\frac{d^2y}{dx^2}+ 6x\frac{dy}{dx}+ 4y= 0\)
That is what is known as an "Euler-type" (because everything is named for Euler!) or "equi-potential" (because the power of x in each term is equal to the order of the derivative). There are typically taught in an introductory Differential Equations course soon after "equations with constant coefficients" because the substitution u= ln(x) (That's what you meant to say, right) converts it to an equation with constant coefficients. That is, I think, what you are trying to do.
With \(\displaystyle u= ln(x)\), \(\displaystyle \frac{du}{dx}= \frac{1}{x}\) so \(\displaystyle \frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}= \frac{1}{x}\frac{dy}{du}\).
And then \(\displaystyle \frac{d^2y}{dx^2}= \frac{d}{dx}\left(\frac{dy}{dx}\right)= \frac{d}{dx}\left(\frac{1}{x}\frac{dy}{du}\right)= \frac{1}{x}\frac{d}{dx}\left(\frac{dy}{du}\right)- \frac{1}{x^2}\frac{dy}{du}= \frac{1}{x^2}\frac{d^2y}{du^2}- \frac{1}{x^2}\frac{dy}{du}\).
Putting those into the equation we have \(\displaystyle \frac{d^2y}{du^2}- \frac{dy}{du}+ 6\frac{dy}{du}+ 4y= \frac{d^2y}{du^2}+ 5\frac{dy}{du}+ 4y= 0\).
That equation has "characteristic equation" \(\displaystyle r^2+ 5r+ 4= (r+ 4)(r+ 1)= 0\) so characteristic roots r= -4 and r= -1. The general solution to this equation is \(\displaystyle y(u)= Ae^{-4u}+ Be^{-u}\). Since u= ln(x), \(\displaystyle y(x)= Ae^{-4ln(x)}+ Be^{-ln(x)}= Ae^{ln(x^{-4}}+ Be^{ln(x^{-1}}= Ax^{-4}+ Bx^{-1}= \frac{A}{x^4}+ \frac{B}{x}\).
(Some texts introduce these equations with the suggestion that you try solutions of the form \(\displaystyle y= x^r\). Had we done that here, since \(\displaystyle y'= rx^{r-1}\) and \(\displaystyle y''= r(r-1)x^{r-2}\) the equation becomes \(\displaystyle r(r-1)x^r+ 6rx^r+ 4x^r= (r^2+ 5r+ 4)x^r= 0\). For this to be true for all x, we must have, as before, \(\displaystyle r^2+ 5r+ 1= (r+ 4)(r+ 1)= 0\) so the two independent solutions are \(\displaystyle x^-4\) and \(\displaystyle x^{-1}\) giving the same solutions. The difficulty is that multiple roots or complex roots of the characteristic equation make this more complicated!)