2
$\begingroup$

I have a first order pde ($P_{14}$ is the unknown function in $a, b, a_1, b_1$): \begin{align} & -\frac{a}{a_1} \frac{\partial P_{14}}{\partial a} - \frac{a b}{a_1 b_1} \frac{\partial P_{14}}{\partial b} + \frac{\partial P_{14}}{\partial a_1} + \frac{a}{a_1} \frac{\partial P_{14}}{\partial b_1} = \frac{-P_{14}}{a_1} + \frac{a^2}{a_1}. \end{align} How to solve this equation? My work: multiply both sides of the equation by $a_1 b_1$, we obtain: \begin{align} & - ab_1 \frac{\partial P_{14}}{\partial a} - ab \frac{\partial P_{14}}{\partial b} + a_1 b_1 \frac{\partial P_{14}}{\partial a_1} + ab_1 \frac{\partial P_{14}}{\partial b_1} = -b_1 P_{14} + a^2 b_1. (1) \end{align}

I think that we need to solve the corresponding homogeneous equation \begin{align} & - ab_1 \frac{\partial P_{14}}{\partial a} - ab \frac{\partial P_{14}}{\partial b} + a_1 b_1 \frac{\partial P_{14}}{\partial a_1} + ab_1 \frac{\partial P_{14}}{\partial b_1} = 0 \end{align} first.

Then we need to solve the equations \begin{align} \frac{da}{-ab_1} = \frac{db}{-ab} = \frac{da_1}{a_1b_1} = \frac{db_1}{ab_1}. \end{align}

Can Maple solve equation (1) directly? What are the solutions of equation (1) (I don't have Maple on my computer)? Thank you very much.

$\endgroup$
1
$\begingroup$

The solution of the equation is \begin{align} P_{14} = -a^2+a F(a_1 a, b_1+a, b b_1), \end{align} where $F(x,y,z)$ is any function in $x,y,z$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.