To solve a non-homogeneous linear PDE $\displaystyle \frac{\partial^2 z}{\partial x^2}+\frac{\partial^2 z}{\partial x \, \partial y}+\frac{\partial z}{\partial y}-z=e^{-x}$

My Attempt: Putting $\displaystyle D=\frac{\partial z}{\partial x}$ and$\displaystyle D’=\frac{\partial z}{\partial y}$, we have $\displaystyle (D^2+DD’+D’-1)z=e^{-x}$


C.F. $\displaystyle \implies z=e^{-x}\phi_1(y) +e^x\phi_2(y-x))$

$$ \mathrm{P.I.}=\frac{1}{f(D,D')}e^{-x}=\frac{1}{(1+1)(D+D'-1)}$$

$$ P.I.=\frac{1}{2(D+D'-1)}e^{-x}$$ (Putting $D=1$ into first factor)

Putting $D=1$ in next factor will get us zero, hence, we move to $$ \frac{1}{(D-mD')}F(x,y)=\int F(x,c-mx)dx=\int e^{-x}dx=-e^{-x}$$

So we finally get, C.F. + P.I. $=\displaystyle e^{-x}\phi_1(y) +e^x\phi_2(y-x))- \frac{e^{-x}}{2}$

The given answer: $$ e^{-x}\phi_1(y) +e^x\phi_2(y-x))- \frac{xe^{-x}}{2}$$

Where does the x come in here as per the given answer? Where am I going wrong?


1 Answer 1


The pde \begin{align} \psi_{xx} + \psi_{xy} + \psi_{y} - \psi = e^{-x} \end{align} can be solved by first making the change \begin{align} \psi(x,y) = f(x,y) - \frac{x}{2} e^{-x} \end{align} which leads to the equation \begin{align} f_{xx} + f_{xy} + f_{y} - f = 0. \end{align} Since the derivative with respect to $y$ is of first order then it is expected to be of an exponential form. This suggests \begin{align} f(x,y) = g(x) e^{-\alpha y}. \end{align} The equation for $g$ is given by \begin{align} g'' - \alpha g' -(\alpha + 1) g = 0 \end{align} and has the solution \begin{align} g(x) = A e^{(\alpha +1)x } + B e^{-x}. \end{align} Combining all the factors together the solution to the pde is \begin{align} \psi(x,y) = A e^{(\alpha + 1)x - \alpha y} + B e^{-x -\alpha y} - \frac{x}{2} e^{-x}. \end{align}

This does have a similar functional form as proposed in the problem, namely, \begin{align} \psi(x,y) = e^{x} \phi(x-y) + e^{-x} \phi(y) - \frac{x}{2} e^{-x} \end{align} where \begin{align} \phi(z) = e^{\alpha z}. \end{align}

  • $\begingroup$ How do we arrive at $\displaystyle \psi(x,y) = f(x,y) - \frac{x}{2} e^{-x}$ ? $\endgroup$
    – square_one
    Commented Jun 27, 2014 at 4:56
  • $\begingroup$ @user148176 $f(x,y)$ solves the homogeneous equation and $x e^{-x}$ is the component that solves the inhomogeneous part. $\endgroup$
    – Leucippus
    Commented Jun 27, 2014 at 5:04
  • $\begingroup$ Thanks.. I have just started to learn this stuff and your method seems new/different from what I am reading. Could you see my attempt and find where I am going wrong? It would also help me wrap my head around "that" method of solving. $\endgroup$
    – square_one
    Commented Jun 27, 2014 at 5:13
  • $\begingroup$ @user148176 Your answer is nearly the same as the one I presented in terms of final result. The only major difference is your $\phi_{1}$ and $\phi_{2}$ should be the same type of function. Other than that your solution appears to be correct. As to the missing $x$ factor somehow the integration method you have used should have the $x$ term as well. Differentiating $x e^{-x}$ leads to the correct form to cancel the $e^{-x}$ term on the right-hand side of the equation. $\endgroup$
    – Leucippus
    Commented Jun 27, 2014 at 5:32
  • $\begingroup$ It seems to me that the problem might be with (D+D'-1), i.e.(non homogeneity). I am applying a rule used for (D-mD'). But I am not sure if its that or did I just make a simple mistake somewhere. $\endgroup$
    – square_one
    Commented Jun 27, 2014 at 5:38

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