# Math Help - x^{2}u_{x}-u_{y}+u=0

1. ## x^{2}u_{x}-u_{y}+u=0

$x^{2}u_{x}-u_{y}+u=0$

$u(x,y)=\varphi(x)\psi(y)$

$x^2\varphi'(x)\psi(y)-\varphi(x)\psi'(y)+\varphi(x)\psi(y)=0$

$x^2\varphi'(x)\psi(y)+\varphi(x)[\psi(y)-\psi'(y)]=0$

$\displaystyle\frac{x^2\varphi'(x)}{\varphi(x)}=-\left[\frac{\psi(y)-\psi'(y)}{\psi(y)}\right]$

$\displaystyle x^2\varphi'(x)-\lambda\varphi(x)=0\Rightarrow x^2m-\lambda=0\Rightarrow m=\frac{\lambda}{x^2}$

$\displaystyle\lambda=\frac{\psi'(y)}{\psi(y)}-1$

$\psi'(y)-\lambda\psi(y)-1=0\Rightarrow n-\lambda-1=0\Rightarrow n=\lambda+1$

$\displaystyle u(x,y;\lambda)=\exp{\left(\frac{\lambda}{x}\right) }*\exp{(y[\lambda+1])}$

Can this method be used with variable coefficients?

Thanks.

2. You need to learn about the method of charactistics. Not everything admits a separation of variables.

3. Originally Posted by dwsmith
$x^{2}u_{x}-u_{y}+u=0$

$u(x,y)=\varphi(x)\psi(y)$

$x^2\varphi'(x)\psi(y)-\varphi(x)\psi'(y)+\varphi(x)\psi(y)=0$

$x^2\varphi'(x)\psi(y)+\varphi(x)[\psi(y)-\psi'(y)]=0$

$\displaystyle\frac{x^2\varphi'(x)}{\varphi(x)}=-\left[\frac{\psi(y)-\psi'(y)}{\psi(y)}\right]$

$\displaystyle x^2\varphi'(x)-\lambda\varphi(x)=0\Rightarrow x^2m-\lambda=0\Rightarrow m=\frac{\lambda}{x^2}$

$\displaystyle\lambda=\frac{\psi'(y)}{\psi(y)}-1$

$\psi'(y)-\lambda\psi(y)-1=0\Rightarrow n-\lambda-1=0\Rightarrow n=\lambda+1$

$\displaystyle u(x,y;\lambda)=\exp{\left(\frac{\lambda}{x}\right) }*\exp{(y[\lambda+1])}$

Can this method be used with variable coefficients?

Thanks.
Dear dwsmith,

No. Suppose we have a differential equation such as, $\displaystyle\varphi'(x)-\lambda\varphi(x)=0$ We take the trail solution as, $\varphi(x)=Ae^{mx}$
Why do we take this exponential function as our trail solution? This is based on the realization that the exponential function keeps its original shape even after differentiation. That is even after taking the derivative you get a the same exponential function with a different coefficient. This is necessary to substract the two functions, $\varphi(x)~and~\varphi'(x)$. The same arguement is valid for any constant coefficient linear differential equation(either partial or ordinary). But when the equation is, $\displaystyle x^2\varphi'(x)-\lambda\varphi(x)=0$ taking the solution as $\varphi(x)=Ae^{mx}$ is not justifiable since the equation is not a constant coefficient one. For more information refer, Linear differential equation - Wikipedia, the free encyclopedia

Hope this helps to clarify matters.

4. Originally Posted by Danny
You need to learn about the method of charactistics. Not everything admits a separation of variables.
Only on page 14 of the book. I am learning as it is presented.

5. Originally Posted by Sudharaka
Dear dwsmith,

No. Suppose we have a differential equation such as, $\displaystyle\varphi'(x)-\lambda\varphi(x)=0$ We take the trail solution as, $\varphi(x)=Ae^{mx}$
Why do we take this exponential function as our trail solution? This is based on the realization that the exponential function keeps its original shape even after differentiation. That is even after taking the derivative you get a the same exponential function with a different coefficient. This is necessary to substract the two functions, $\varphi(x)~and~\varphi'(x)$. The same arguement is valid for any constant coefficient linear differential equation(either partial or ordinary). But when the equation is, $\displaystyle x^2\varphi'(x)-\lambda\varphi(x)=0$ taking the solution as $\varphi(x)=Ae^{mx}$ is not justifiable since the equation is not a constant coefficient one. For more information refer, Linear differential equation - Wikipedia, the free encyclopedia

Hope this helps to clarify matters.
How am I suppose to approach this problem then? My book hasn't covered it yet.

6. $\displaystyle\frac{x^2\varphi'(x)}{\varphi(x)}=-\left[\frac{\psi(y)-\psi'(y)}{\psi(y)}\right]$

$x^2\varphi'(x)-\lambda\varphi(x)=0$

$\varphi(x)=x^m$

$x^2mx^{m-1}-x^m\lambda=0\Rightarrow x^m[x^2mx^{-1}-\lambda]=0\Rightarrow x^m[xm-\lambda]=0$

Can I have an x in my m solution?

$\displaystyle m=\frac{\lambda}{x}\mbox{?}$

7. Nope. To solve this separate variables

$\dfrac{d \varphi}{\varphi} = \dfrac{\lambda}{x^2}$ and integrate.

8. $\displaystyle\int\frac{\varphi'(x)}{\varphi(x)}=\i nt\frac{\lambda}{x^2}\varphi'(x)\Rightarrow \ln{(\varphi(x))}=\frac{-\lambda}{x}+c\Rightarrow \varphi(x)=C_2\exp{\left(\frac{-\lambda}{x}\right)}$

$\displaystyle u(x,y)=C_2\exp{\left(\frac{-\lambda}{x}\right)}C_3\exp{(y(\lambda+1))}=C_4\exp {\left(\frac{-\lambda}{x}\right)}\exp{(y(\lambda+1))}$

Correct?

9. Originally Posted by dwsmith
$x^{2}u_{x}-u_{y}+u=0$

$u(x,y)=\varphi(x)\psi(y)$

$x^2\varphi'(x)\psi(y)-\varphi(x)\psi'(y)+\varphi(x)\psi(y)=0$

$x^2\varphi'(x)\psi(y)+\varphi(x)[\psi(y)-\psi'(y)]=0$

$\displaystyle\frac{x^2\varphi'(x)}{\varphi(x)}=-\left[\frac{\psi(y)-\psi'(y)}{\psi(y)}\right]$

$\displaystyle x^2\varphi'(x)-\lambda\varphi(x)=0\Rightarrow x^2m-\lambda=0\Rightarrow m=\frac{\lambda}{x^2}$

$\displaystyle\lambda=\frac{\psi'(y)}{\psi(y)}-1$

$\psi'(y)-\lambda\psi(y)-1=0\Rightarrow n-\lambda-1=0\Rightarrow n=\lambda+1$

$\displaystyle u(x,y;\lambda)=\exp{\left(\frac{\lambda}{x}\right) }*\exp{(y[\lambda+1])}$

Can this method be used with variable coefficients?

Thanks.
Dear dwsmith,

There is a mistake, but somehow you have obtained the correct answer. The correct method is,

$\displaystyle\lambda=\frac{\psi'(y)}{\psi(y)}-1\Rightarrow \psi'(y)-\lambda\psi(y)-\psi(y)=0\Rightarrow n-\lambda-1=0\Rightarrow n=\lambda+1$

10. I have it written down on paper correctly though

11. Originally Posted by dwsmith
$\displaystyle\int\frac{\varphi'(x)}{\varphi(x)}=\i nt\frac{\lambda}{x^2}\varphi'(x)\Rightarrow \ln{(\varphi(x))}=\frac{-\lambda}{x}+c\Rightarrow \varphi(x)=C_2\exp{\left(\frac{-\lambda}{x}\right)}$

$\displaystyle u(x,y)=C_2\exp{\left(\frac{-\lambda}{x}\right)}C_3\exp{(y(\lambda+1))}=C_4\exp {\left(\frac{-\lambda}{x}\right)}\exp{(y(\lambda+1))}$

Correct?
Dear dwsmith,

Well there are some problems. The integration should be, $\displaystyle\int\frac{d(\varphi(x))}{\varphi(x)}= \int\frac{\lambda}{x^2}dx$ and you have forgotten the absolute value of the natural logarithm.

$\ln{\mid\varphi(x)\mid}=\dfrac{-\lambda}{x}+c$

This wont change the answer but still it matters.