# Find the general solution of the given PDE

• chwala
In summary, we can solve the given PDE by transforming to the variables (\zeta,\eta) given by x = -2\zeta + \frac12\eta,\qquad y = 3\zeta + \frac14\eta and then solving the resulting ODE. The solution is given by u = f(\eta)e^{-5\zeta} + \frac{1}{18}e^{4\zeta} where \zeta = \frac14(x+2y) and \eta = 3x+2y.

#### chwala

Gold Member
Homework Statement
This is a question set by myself;

Find the general solution of

##-4u_x+6u_y+10u=e^{x+2y}##
Relevant Equations
Method of characteristic.
My take;

##ξ=-4x+6y## and ##η=6x+4y##

it follows that,

##52u_ξ +10u=e^{x+2y}##

for the homogenous part; we shall have the general solution;

$$u_h=e^{\frac{-5}{26} ξ} f{η }$$

now we note that

$$e^{x+2y}=e^{\frac{8ξ+η}{26}}$$

that is from solving the simultaneous equation;

##ξ=-4x+6y## and ##η=6x+4y## where ##x=\dfrac{12ξ-18η}{156}## and ##y=\dfrac{3ξ+2η}{26}##

for the inhomogenous part, we shall have;

$$u_p = e^{\frac{-5}{26}ξ}\int \frac{e^{\frac{8ξ+η}{26}}⋅e^{\frac{5}{26}ξ}}{52} dξ$$$$u_p = \frac{e^{\frac{13ξ+η}{26}}⋅e^{\frac{-5}{26}ξ}}{26}$$

$$u(ξ,η)=u_h+u_p=e^{\frac{-5}{26} ξ} \left[ f(η)+\frac{e^{\frac{13ξ+η}{26}}}{26}\right]$$

$$u(x,y)= e^{\frac{-5}{26} (-4x+6y)} \left[ f(6x+4y)+\frac{1}{26} {e^{\frac{-46x+82y}{26}}}\right]$$your insight is welcome or alternative approach to this problem.

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First divide by 2: $$-2 u_x + 3u_y + 5u = \tfrac12 e^{x+ 2y}.$$ We want to change to variables $(\zeta, \eta)$ with $x_\zeta = -2$ and $y_\zeta = 3$ so that $$-2u_x + 3u_y = u_xx_\zeta + u_yy_\zeta = u_\zeta.$$ Note that you did the opposite, which was to set $$-4 u_x + 6u_y = u_x\xi_x + u_y \xi_y$$ where the right hand side is not the multivariate chain rule for $u_\xi$.

We therefore set $$\begin{split} x &= -2\zeta + a\eta \\ y &= 3\zeta + b\eta \end{split}$$ with $a$ and $b$ chosen such that $$x + 2y = 4\zeta + (a + 2b)\eta = 4\zeta.$$ Any choice of $b$ other than $b = 0$ gives us a 1-1 transformation, but the choice $b = -\frac14$ gives the convenient $\eta = 3x + 2y$, and from the above we have $\zeta = \frac14(x+2y)$. The PDE is then reduced to $$u_\zeta + 5u = \tfrac12e^{4\zeta}$$ which can be solved by an integrating factor.

SammyS
@pasmith I'll need to go through your approach...cheers man! Just curious are we going to get same solutions? Your integrating factor seems different just by looking at it...

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I get $$u = g(\eta')e^{-5\zeta} + \tfrac{1}{18}e^{4\zeta}.$$

I think you have made an error: If $\xi = -4x + 6y$ and $\eta = 6x + 4y$ then $$x = (3\eta - 2\xi)/26,\qquad y = (3\xi + 2\eta)/26.$$ Your result of $$x = \frac{12\xi - 18\eta}{156} = \frac{2\xi - 3\eta}{26}$$ is out by a factor of -1, so your calculation of $x + 2y$ is incorrect; you should have $$x + 2y = \frac{7\eta + 4\xi}{26}.$$ Note that in terms of my $\zeta, \eta'$ you have $$\begin{split} \eta &= 2\eta' \\ \xi &= 26\zeta - \frac{7}{2}\eta'.\end{split}$$ Now solving $$u_\xi + \frac{5}{26}u = \frac{1}{52}e^{(7\eta +4\xi)/26}$$ gives $$\begin{split} u &= f(\eta)e^{-5\xi/26} + \frac{1}{52}\frac{26}{9}e^{(7\eta+4\xi)/26} \\ &= f(\eta)e^{-5\xi/26} + \frac{1}{18}e^{(7\eta + 4\xi)/26}.\end{split}$$ We agree on the second term, since $$4\zeta = x + 2y = \frac{7\eta + 4\xi}{26}.$$ For the first term, we have $$f(\eta)e^{-5\xi/26} = \overbrace{f(2\eta')e^{35\eta'/52}}^{\equiv g(\eta')}e^{-5\zeta}$$ as required.

I suspect you will agree that the numbers in my solution are simpler than in yours.

chwala

## 1. What is a PDE?

A PDE stands for a partial differential equation, which is an equation that involves partial derivatives of a function with respect to multiple independent variables.

## 2. What is a general solution of a PDE?

A general solution of a PDE is a solution that satisfies the equation for all possible values of the independent variables. It contains an arbitrary constant or a set of arbitrary constants that can take on any value.

## 3. How do you find the general solution of a PDE?

To find the general solution of a PDE, you need to use techniques such as separation of variables, substitution, or the method of characteristics. These methods involve manipulating the equation to isolate the dependent variable and then solving for it.

## 4. What is the difference between a general solution and a particular solution?

A general solution contains arbitrary constants and represents all possible solutions to the PDE, while a particular solution is a specific solution that satisfies the equation for given values of the independent variables. A particular solution can be obtained by assigning specific values to the arbitrary constants in the general solution.

## 5. Are there any limitations to finding the general solution of a PDE?

Yes, there are limitations to finding the general solution of a PDE. In some cases, it may not be possible to find a general solution using known methods. Additionally, the general solution may not always be unique, and there may be an infinite number of solutions that satisfy the PDE.