Question

Find the singular solution of Partial differential equation z=p+q+2√pq

Answer 1

Step-by-step explanation:

Charpit’s auxiliary equations are

dp∂f∂x+p∂f∂z=dq∂f∂y+q∂f∂z=dz−p∂f∂p−q∂f∂q=dx−p∂f∂p=dy−q∂f∂q=dF0

After getting all the required values, we have

dpp=dqq=dz2pq=dxq=dyp=dF0

Taking second and fourth factors, we get

dqq=dxq⟹dq=dx

Integrating, we get

q=x+a

After putting this value in the given equation, we get

p=zx+a

Now dz=pdx+qdy gives

⟹dz=zx+adx+(x+a)dy(x+a)dz−zdx(x+a)2=dy

On integration, we get

zx+a=y+b

i.e z=(x+a)(y+b)