Question : Find the solution of pde 3d^2z/dx^2+2d^2z/dxdy-5d^2z/dy^2=0
Solution :
Given PDE [3\frac{\partial^2z}{\partial x^2}+2\frac{\partial^2z}{\partial x\partial y}-5\frac{\partial^2z}{\partial y^2}=0 ] Is a Homogeneous Linear PDE with constant coefficients. To solve such equations we process as follows Step 1: Writing Auxiliary equation Substitute [\frac{\partial z}{\partial x}=m,\ \frac{\partial z}{\partial x}=1] We get auxiliary equation as [3m^2+2m-5=0] Step2 : Finding roots of Auxiliary equation [3m^2+2m-5=0] [=>3m^2+5m-3m-5=0] [=>m\left(3m+5\right)-1\left(3m+5\right)=0] [=>\left(3m+5\right)\left(m-1\right)=0] [=>m=1,\ -\frac{5}{3}] These are real and distinct roots. Step 3 : Writing General solution of the PDE [z=c\phi\left(y+1x\right)+d\phi\left(y-\frac{5}{3}x\right)] Where c and d are arbitrary constat and \phi is an arbitrary function. Question : 2 Solution : Given PDE r+10s+25t =0 we know the notation Given PDE [r=\frac{\partial^2z}{\partial x^2},\ s=\frac{\partial^2z}{\partial x\partial y}\ ,\ t=\frac{\partial^2z}{\partial y^2} ] Hence the given PDE becomes [\frac{\partial^2z}{\partial x^2}+10\frac{\partial^2z}{\partial x\partial y}+25\frac{\partial^2z}{\partial y^2}=0] Which Is also a Homogeneous Linear PDE with constant coefficients. To solve this PDE we process as above We get auxiliary equation as [m^2+10m+25=0] [=>\left(m+5\right)^2=0] [=>m=-5,\ -5] We obtained real and and equal roots. Therefore General solution of the given PDE is as follows [z=(A+Bx)\phi\left(y-5x\right)] Where A and B are arbitrary constants and \phi is an arbitrary function.