Answer:
Following are the solution to this question:
Step-by-step explanation:
differentail equation:
[tex]yu_{xy} + u = 0..........(1)[/tex]
when [tex]u(x,y) = X(x)Y(y)[/tex]
[tex]u_x = \frac{\partial u}{\partial x} X'Y' \\\\ u_{xy} = \frac{\partial^2 u }{ \partial x\ \partial y}= X'Y'[/tex]
by substituting the value we get:
[tex]\to \int \frac{Y'}{Y} = -\lambda \int \frac{1}{y} dy\\\\[/tex]
by solving the value:
[tex]\to Y=c_3 y^{-\lambda}\\\\\therefore\ \ c_4=c_2C_3\\\\\to u(x,y) = c_4 e^{\frac{x}{\lambda}} y^{-\lambda}[/tex]
The solution of the given partial differential equation will be
[tex]u(x,y)=c_4e^\frac{x}{\lambda}y^{-\lambda}[/tex]
The partial derivative is a way to find the slope in either the x or y direction, at the point indicated.
[tex]yu_{xy}+u=0..............................(1)[/tex]
Now if [tex]u(x,y)=X(x)Y(y)[/tex]
[tex]u_x=\dfrac{\partial u}{\partial x} X'Y'[/tex]
[tex]u_{xy}=\dfrac{\partial^2u}{\partial x\ \partial y}=X'Y'[/tex]
by substituting the value, we get:
[tex]\int \dfrac{Y'}{Y}=-\lambda\int\dfrac{1}{y}dy[/tex]
by solving the value:
[tex]Y=c_3y^{-\lambda}[/tex]
[tex]c_4=c_2c_3[/tex]
[tex]u(x,y)=c_4e^\frac{x}{\lambda}y^{-\lambda}[/tex]
Hence the solution of the given partial differential equation will be
[tex]u(x,y)=c_4e^\frac{x}{\lambda}y^{-\lambda}[/tex]
To know more about partial differential equation follow
https://brainly.com/question/2293382
