Is this the correct general solution of the given PDE?

[Math100]

Homework Statement

Find the general solution of the partial differential equation $u_{xx}+6u_{x}+9u=0$.

Relevant Equations

None.

Find the general solution of the partial differential equation $u_{xx}+6u_{x}+9u=0$.

Here's my work:

Let $u(x, y)=e^{rx}$.
Then $u_{x}(x, y)=re^{rx}$ and we get that $u_{xx}(x, y)=r^2e^{rx}$.
By using direct substitution of $u=e^{rx}, u_{x}=re^{rx}$ and $u_{xx}=r^2e^{rx}$,
we have that $u_{xx}+6u_{x}+9u=0\implies r^2e^{rx}+6re^{rx}+9e^{rx}=0\implies e^{rx}(r^2+6r+9)=0\implies r^2+6r+9=0\implies (r+3)^2=0$.
Thus, $r=-3$.
Therefore, the general solution of the partial differential equation $u_{xx}+6u_{x}+9u=0$ is
$u(x, y)=A(y)e^{-3x}+B(y)\cdot xe^{-3x}$ where $A, B$ are arbitrary functions of $y$.

Above is my work with the answer. May someone please check/verify to see if everything is correct? Also, I want to know where does the substitution of $u(x, y)=e^{rx}$ comes from, although I was just told to use it without any definition, lemma, theorem, etc. As for the characteristic equations with repeated roots like the given partial differential equation above, will its general solution always of the form $u(x, y)=A(y)e^{rx}+B(y)\cdot xe^{rx}$ where $A, B$ are arbitrary functions of $y$?

 

[Mark44]

Your initial assumption about the possible solution is that $u(x, y) = e^{rx}$. It ignores the fact that the solution should also depend on y. This work would be appropriate for an ODE such as y'' + 6y' + 9y = 0, where y is a function of a single variable. For such a problem, a place to start is to assume that a solution is $y = e^{rx}$. There doesn't need to be a definition, theorem, or lemma -- it's just a technique that may or may not produce results.

It's been a really long time since I had to solve PDEs, so I'll have to dig out my textbook to see if your approach is reasonable or not.

 

[Chestermiller]

If your solution satisfies the differential equation, then it must be correct. (assuming y is the only other independent variable involved).

 

[martinbn]

Your initial assumption about the possible solution is that $u(x, y) = e^{rx}$. It ignores the fact that the solution should also depend on y. This work would be appropriate for an ODE such as y'' + 6y' + 9y = 0, where y is a function of a single variable. For such a problem, a place to start is to assume that a solution is $y = e^{rx}$. There doesn't need to be a definition, theorem, or lemma -- it's just a technique that may or may not produce results.

It's been a really long time since I had to solve PDEs, so I'll have to dig out my textbook to see if your approach is reasonable or not.

I think the idea is that if $u(x,y)$ is any solution of the equations, and if you fix $y$, then the function $z=u(x,y)$ in $x$ is a solution of $z''+6z'+z=0$ so it has to be of the form $z(x)=Ae^{-3x}+Bxe^{-3x}$. For different $y's$ the constants $A$ and $B$ will be different i.e. they are functions of $y$. So $u(x,y)= A(y)e^{-3x}+B(y)xe^{-3x}$

 

[Math100]

Your initial assumption about the possible solution is that $u(x, y) = e^{rx}$. It ignores the fact that the solution should also depend on y. This work would be appropriate for an ODE such as y'' + 6y' + 9y = 0, where y is a function of a single variable. For such a problem, a place to start is to assume that a solution is $y = e^{rx}$. There doesn't need to be a definition, theorem, or lemma -- it's just a technique that may or may not produce results.

It's been a really long time since I had to solve PDEs, so I'll have to dig out my textbook to see if your approach is reasonable or not.

Now I think that my assumption could be wrong then, because like you said above, we shouldn't pretend to solve such problems like we used to approach ODE problems since this is a PDE instead of an ODE. So $u(x, y)$ should equal to something else, rather than just $u(x, y)=e^{rx}$. But what should $u(x, y)$ be? That's the problem. Also, do we really have to form a guess for such $u(x, y)$ to be something? In addition, I put this PDE on my graphing calculator with the following command: deSolve(diff(u(x, y), x, 2)+6*diff(u(x, y), x)+9*u(x, y)=0, u(x, y)) for $u_{xx}+6u_{x}+9u=0$ but my graphing calculator says "Too few arguments" after I hit the button 'Enter' for answer/result. It'd be nice to find the final answer on graphing calculator so at least I can check the answer by myself but that didn't help either.

 

[mad mathematician]

Homework Statement: Find the general solution of the partial differential equation $u_{xx}+6u_{x}+9u=0$.
Relevant Equations: None.

Find the general solution of the partial differential equation $u_{xx}+6u_{x}+9u=0$.

Here's my work:

Let $u(x, y)=e^{rx}$.
Then $u_{x}(x, y)=re^{rx}$ and we get that $u_{xx}(x, y)=r^2e^{rx}$.
By using direct substitution of $u=e^{rx}, u_{x}=re^{rx}$ and $u_{xx}=r^2e^{rx}$,
we have that $u_{xx}+6u_{x}+9u=0\implies r^2e^{rx}+6re^{rx}+9e^{rx}=0\implies e^{rx}(r^2+6r+9)=0\implies r^2+6r+9=0\implies (r+3)^2=0$.
Thus, $r=-3$.
Therefore, the general solution of the partial differential equation $u_{xx}+6u_{x}+9u=0$ is
$u(x, y)=A(y)e^{-3x}+B(y)\cdot xe^{-3x}$ where $A, B$ are arbitrary functions of $y$.

Above is my work with the answer. May someone please check/verify to see if everything is correct? Also, I want to know where does the substitution of $u(x, y)=e^{rx}$ comes from, although I was just told to use it without any definition, lemma, theorem, etc. As for the characteristic equations with repeated roots like the given partial differential equation above, will its general solution always of the form $u(x, y)=A(y)e^{rx}+B(y)\cdot xe^{rx}$ where $A, B$ are arbitrary functions of $y$?

it looks like the solution is $u(x,y)=g(y)e^{rx}$. And as you know you get $r=-3$ with multiplicity 2; so the general solution is: $u(x,y)=A(y)e^{-3x}+B(y)xe^{-3x}$.

 

[mad mathematician]

An interesting variation may be this one: $u_{xx}+6u_x+9u_y=0$, which can be solved by the method of characteristics: $dx/ds=6, dy/ds=9, du/ds=-u_{xx}$, $x(s)=6s, y(s)=9s$ plug and get: $du/ds=-1/36 d^2u/ds^2$ which can be solved by repeated integration to get: $du/ds+36u=c$.
Easy peasy from here.

edit: it seems the pde I suggested is the heat pde in disguise;
you can't solve it in the way I suggested above.

 

[Gavran]

Now I think that my assumption could be wrong then, because like you said above, we shouldn't pretend to solve such problems like we used to approach ODE problems since this is a PDE instead of an ODE. So $u(x, y)$ should equal to something else, rather than just $u(x, y)=e^{rx}$. But what should $u(x, y)$ be? That's the problem. Also, do we really have to form a guess for such $u(x, y)$ to be something? In addition, I put this PDE on my graphing calculator with the following command: deSolve(diff(u(x, y), x, 2)+6*diff(u(x, y), x)+9*u(x, y)=0, u(x, y)) for $u_{xx}+6u_{x}+9u=0$ but my graphing calculator says "Too few arguments" after I hit the button 'Enter' for answer/result. It'd be nice to find the final answer on graphing calculator so at least I can check the answer by myself but that didn't help either.

Your method and your result in the original post are correct because every partial differential equation that only involves derivatives with respect to one variable can be treated as an ordinary differential equation. If a partial differential equation involves derivatives with respect to more than one variable, your method does not work.

 

[wrobel]

An interesting variation may be this one: $u_{xx}+6u_x+9u_y=0$, which can be solved by the method of characteristics: $dx/ds=6, dy/ds=9, du/ds=-u_{xx}$, $x(s)=6s, y(s)=9s$ plug and get: $du/ds=-1/36 d^2u/ds^2$ which can be solved by repeated integration to get: $du/ds+36u=c$.
Easy peasy from here.

edit: it seems the pde I suggested is the heat pde in disguise;
you can't solve it in the way I suggested above.

yes and you can reduce it to the standard form by the substitution $u=e^{-3x+y}w(x,y)$

 

[Math100]

Your method and your result in the original post are correct because every partial differential equation that only involves derivatives with respect to one variable can be treated as an ordinary differential equation. If a partial differential equation involves derivatives with respect to more than one variable, your method does not work.

By checking/verifying/confirming my final answer/solution for $u(x, y)=[A(y)+B(y)\cdot x]e^{-3x}$, I've obtained $u_{x}(x, y)=e^{-3x}[-3A(y)+B(y)(1-3x)], u_{xx}=e^{-3x}[9A(y)+9x\cdot B(y)-6B(y)]$.
Thus, after using/applying the direct substitution method, I got/obtained that
$u_{xx}+6u_{x}+9u=0\implies e^{-3x}[9A(y)+9x\cdot B(y)-6B(y)]+6e^{-3x}[-3A(y)+B(y)(1-3x)]+ 9e^{-3x}[A(y)+B(y)\cdot x]=18e^{-3x}A(y)+18xe^{-3x}\cdot B(y)-18e^{-3x}A(y)-6e^{-3x}B(y)+ 6e^{-3x}B(y)-18xe^{-3x}\cdot B(y)=0$.
This implies that the final answer/solution $u(x, y)=A(y)e^{-3x}+B(y)\cdot xe^{-3x}$ where $A, B$ are arbitrary functions of $y$ is correct. However, may someone please explain why, how, and where does $u(x, y)=e^{rx}$ for such PDE problems? Does someone also know the correct command for putting these type(s) of PDEs into a graphing calculator and getting the exact answer/solution for verification/confirmation?

 

[wrobel]

this is an ODE not PDE

 

[Gavran]

However, may someone please explain why, how, and where does $u(x, y)=e^{rx}$ for such PDE problems?

https://en.wikipedia.org/wiki/Linea...mogeneous_equation_with_constant_coefficients

Does someone also know the correct command for putting these type(s) of PDEs into a graphing calculator and getting the exact answer/solution for verification/confirmation?

https://www.wolframalpha.com/input?...l[u\(40)x\(44)y\(41),x]+9u\(40)x\(44)y\(41)=0

 

[Math100]

https://en.wikipedia.org/wiki/Linea...mogeneous_equation_with_constant_coefficients


https://www.wolframalpha.com/input?i2d=true&i=Partial[u\(40)x\(44)y\(41),{x,2}]+6Partial[u\(40)x\(44)y\(41),x]+9u\(40)x\(44)y\(41)=0

Thank you very much for the sources and help!