I have a question
I want to solve the following $xu_x+yu_y= xy\sin(xy)$ and $2u_x+3u_y= 4x-9y^2$
These are linear PDEs. So the solution would be a sum of the homogeneous solution and particular solution. I just dont know how to get the particular solutions. I'm not even sure what to guess. What would the particular solutions be?
Sometimes by inspection one can guess the form of a particular solution. But this is not often the case. Most likely one have to solve the PDE thanks to the method of characteristics for example.
First equation :$$xu_x+yu_y= xy\sin(xy)\tag 1$$
The term on right side with $xy$ into it draw to look for a particular solution on the form $$u=f(X)\quad\text{with}\quad X=xy$$ $u_x=yf'(X)\quad$ and $\quad u_y=xf'(X)$ $$xu_x+yu_y=2Xf'(X) =X\sin(X)$$ $$2f'(X)=\sin(X)$$ $$f(X)=-\frac12 \cos(X)$$ Doesn't mater the integration constant since we look for any one particular solution. $$u=-\frac12 \cos(xy)\quad\text{is a particular solution}$$ The solution of the homogeneous PDE is $u=F\left(\frac{y}{x}\right)$ with arbitrary function $F$.
So, the general solution of the PDE $(1)$ is : $$u(x,y)=F\left(\frac{y}{x}\right)-\frac12 \cos(xy)$$
Solving the PDE with the method of characteristics :
The Charpit-Lagrange system of characteristic ODEs is : $$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{xy\sin(xy)}$$ A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{y}$ : $$\frac{y}{x}=c_1$$ A second characteristic equation comes from $\frac{dx}{x}=\frac{dy}{y}=\frac{ydx+xdy}{yx+xy}=\frac{d(xy)}{2xy}=\frac{du}{xy\sin(xy)}$ $$u+\frac12\cos(xy)=c_2$$ The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ is : $$u+\frac12\cos(xy)=F\left(\frac{y}{x}\right)$$ $$u(x,y)=F\left(\frac{y}{x}\right)-\frac12 \cos(xy)$$
Second equation : $$2u_x+3u_y= 4x-9y^2\tag 2$$ By inspection. We try a particular solution with separated variables : $$u=f(x)+g(y)$$ $u_x=f'(x)\quad$ and $\quad u_y=g'(y)$ $$2u_x+3u_y=2f'(x)+3g'(y)=4x-9y^2\quad\implies\quad f'(x)=2x\quad\text{and}\quad g'(y)=-3y^2$$ $$f(x)=x^2\quad\text{and}\quad g(y)=y^3$$ Doesn't mater the integration constant since we look for any one particular solution.
$$u=x^2-y^3\quad\text{is a particular solution}$$
The solution of the homogeneous PDE is $u=F\left(\frac{y}{3}-\frac{x}{2}\right)$ with arbitrary function $F$.
So, the general solution of the PDE $(2)$ is : $$u(x,y)=F\left(\frac{y}{3}-\frac{x}{2}\right)+x^2-y^3$$
Solving the PDE with the method of characteristics :
The Charpit-Lagrange system of characteristic ODEs is : $$\frac{dx}{2}=\frac{dy}{3}=\frac{du}{4x-9y^2}$$ A first characteristic equation comes from solving $\frac{dx}{2}=\frac{dy}{3}$ : $$\frac{y}{3}-\frac{x}{2}=c_1$$ A second characteristic equation comes from $\frac{dx}{x}=\frac{dy}{y}=\frac{2xdx-3y^2dy}{2(2x)-3(3y^2)}=\frac{du}{4x-9y^2}$
$$u-x^2+y^3=c_2$$ The general solution of the PDE expressed on implicit form $c_2=F(c_1)$ is : $$u-x^2+y^3=F\left(\frac{y}{3}-\frac{x}{2}\right)$$ $$u(x,y)=F\left(\frac{y}{3}-\frac{x}{2}\right)+x^2-y^3$$
