Linked Questions

0
votes
2answers
89 views

equation $u_x+u=e^{2y+x}$ (part of the solution to $u_x+u_y+u=e^{x+2y}$) [duplicate]

I solved/analyzed the below PDE $$\left\{\begin{matrix} u_x+u_y+u=e^{x+2y}\\ u(x,0)=0 \end{matrix}\right.$$ and have a question to the one of the steps involving the integration, see below Using ...
1
vote
1answer
91 views

Solve: $U_x+U_y+U=e^{x+2y}$ with $U(x,0)=0$ [duplicate]

So, I've been tasked with solving the above. I've gone up to the general solution which wasn't too tricky: $$U(x,y)=e^{-x}(\frac{R_1e^{4x}}{4}+f(y-x))$$ However, it gets a bit strange for me when I'...
2
votes
2answers
63 views

I am trying to Solve a linear non-homogeneous PDE [duplicate]

I am trying to solve $$u_x + u_y + u = e^{x+2y} $$ I started this problem by using the coordinate method. I set $$t = x+y$$ $$p = x-y$$(Skipping a couple of steps) I got to $$u_t + \frac{1}{2}u = \...
0
votes
1answer
40 views

Simple 1st order PDE [duplicate]

Solve $u_x+u_y+u=e^{x+2y}$ with $u(x,0)=0$ I try to let $x'=x+y, y'=x-y$ and reduced to $2u_{x'}+u=e^{0.5(3x'-y)}$ How to proceed to the next step? Any other methods to solve? Thank you!
0
votes
0answers
28 views

Partial Differential Equation equal to a function [duplicate]

Hi the question I have to solve is: $u_x + u_y + u = e^{x+2y} \ $ where $ u(x,0) = 0$ First, I tried to solve the question of the form: $$ u_x + u_y + u = 0$$ We know that $\dot x = 1 \ $ and $\...
0
votes
1answer
351 views

Finding particular solutions to PDEs (transport equation)

Suppose I have the following PDEs: (for $u(t,x)$) (1) $u_{t}-u_{x}=-x$ (2) $u_{t}+2u_{x}=1$ (3) $u_{t}+u_{x}+u=e^{x+3t}$ (4) $2u_{t}+u_{x}=sin(x-t)$ For all equations, it is easy to find the ...
1
vote
2answers
86 views

Solve $u_x+u_y+u = e^{x+2y}$ with the condition $u(x,0) = 0$

I did read the answer to here, but my approach is a bit different and would like some suggestions. Solve $u_x + u_y + u = e^{x+2y}$ One checks that $u = \frac{e^{x+2y}}{4}$ is a special solution to $...
1
vote
1answer
151 views

Solve the PDE by the method of characteristics.

I am trying to figure out where my solution went wrong. I am off by a factor of two. $$ u_x + u_y + u = e^{x+2y}$$ I first found that the characteristic curves are determined by $$\frac{dy}{dx} = 1 \...
0
votes
1answer
98 views

Solving a first order PDE with initial value condition by change of variable.

I want to solve the pde $u_x+u_y+u=e^{x+2y}$ with ivc $u(x,0)=0$, which is a specific case of the general pde $au_x+bu_y+cu=g(x,y)$, where $a,b,c\in \mathbb{R}$. Consider the change of coordinates $\...
2
votes
1answer
66 views

Solving an Inhomogeneous $1$st Order PDE using Method of Characteristics

I wanna solve the equation $$u_x+u_y+u=\exp(x+2y), \quad u(x,0) = 0$$ I have just learned method of characteristics. But I don't know how to deal with $u$ term and inhomogeneous term simultaneously. ...