Particular solutions of 3 linear PDEs with nonstandart right-hand side

**bkarpuz** · January 16th 2012, 11:47 PM

Dear MHF members,

I have the following 3 PDEs, whose solutions I know but I am looking for nice explanary solutions. Please help me.

$(D_{x}^{2}+D_{x}D_{y}-2D_{y}^{2})z=8\big(\sec(2x+y)\big)^{2}\tan(2x+y)$ [Solution: $z_{p}=\tan(2x+y)$ ]
$(D_{x}^{2}+D_{x}D_{y}-D_{x}-D_{y})z=1/y^{2}$ [Solution: $z_{p}=1/y$ ]
$(D_{x}^{2}-D_{y}^{2}-D_{x}+D_{y})z=(x+1)/x^{2}$ [Solution: $z_{p}=-\ln(x)$ ]

Thanks.
bkarpuz

**Danny** · January 17th 2012, 04:51 AM

What do you mean by "nice explanary solutions"? Do you mean how they came up with $z_p$ ? If so, I think I can explain this.

**bkarpuz** · January 17th 2012, 05:00 AM

Originally Posted by Danny

What do you mean by "nice explanary solutions"? Do you mean how they came up with $z_p$ ? If so, I think I can explain this.

Exactly!

**Danny** · January 17th 2012, 05:22 AM

The second and third are rather straight forward. For the second, the RHS is only a function of y so sek a particular solution in the form $z_p = f(y).$ You'll get an ODE for $f$ (and it's linear!) Similarly for the third. For the first, you'll notice that the RHS is a function of $2x+y$ so seek $z_p = f(2x+y).$ Since the LHS of the PDEs are independent of both independent variables, this method works.

Hope it helps.

**bkarpuz** · January 17th 2012, 06:21 AM

Originally Posted by Danny

The second and third are rather straight forward. For the second, the RHS is only a function of y so sek a particular solution in the form $z_p = f(y).$ You'll get an ODE for $f$ (and it's linear!) Similarly for the third. For the first, you'll notice that the RHS is a function of $2x+y$ so seek $z_p = f(2x+y).$ Since the LHS of the PDEs are independent of both independent variables, this method works.

Hope it helps.

For the first one, I also noticed that $\frac{\mathrm{d}^{2}\tan(u)}{\mathrm{d}u^{2}}=\big (\sec(u)\big)^{2}\tan(u)$ as the left-hand side is involves only 2nd order derivatives.
Dor the 2nd and the 3rd ones, I rewrote left-hand side as $(D_{x}-1)(D_{x}+D_{y})$ and $D_{x}(D_{x}-1)-D_{y}(D_{y}-1)$ or $(D_{x}+D_{y}-1)(D_{x}-D_{y})$ but I could not see any nice things here...

Is it possible to make change of variables to reduce the equation into first-order equation as in the ordinary case?
Thanks...

**Danny** · January 17th 2012, 06:35 AM

As you've said, for the first, you could made a change of variables where u = 2x + y and pick another independent variable.

For the second and third, since you've noticed that the operators factor, you can introduced a new variable as to make your problem two linear first order PDEs. For example, for the second one, if you let

$w = z_x + z_y$ then $w_x - w = \frac{1}{y^2}$

Solve the second, then use this in the first.

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