Question

(x2-y2)pq-xy(p2-q2)=1 solve partial diferentail equation

Answer 1

Answer:

z =a/2 ln(x^2 + y^2) + 1/a tan−1(y/x)+b

Step-by-step explanation:

Charpit’s auxiliary equations are

dp/fx + pfz=dq/fy + qfz=dz/−pfp − qfq=dx/−fp=dy/−fq

i.e.

dp/2pqx − y(p^2 − q^2) = dq/−2pqy − x(p^2 − q^2) = dx/−(x^2 − y^2)q + 2pqxy dy/−(x^2 − y^2)p − 2pqxy = dz/−p(x^2q − y^2q − 2pxy) − q(px^2 − y^2p+ 2xyq)

Using x, y, p, q, 0 as multiplier, then each fraction is

=

xdp + ydq + pdx + qdy = 0

d(xp + qy)=0

xp + yq = a ⇒

p =a − qy/x

Putting this value of p in given PDE, and solving for q, we get

q =a^2y + x/a(x^2 + y^2)

Now substituting the values of p and q in the equation

dz = pdx + qdy

we get

dz ={(a^2x − y)dx + (a^2y + x)dy}/a(x^2 + y^2)

or

dz = axdx + ydy/(x^2 + y^2)+xdy − ydx/a(x2 + y2)

Integrating above equation, we get the required complete integral as following

z =a/2 ln(x^2 + y^2) + 1/a tan−1(y/x)+b

Answer 2

Solving PDEs analytically is generally based on finding a change of variable to transform the equation into something soluble or on finding an integral form of the solution. a ∂u ∂x + b ∂u ∂y = c. dy dx = b a , and ξ(x, y) independent (usually ξ = x) to transform the PDE into an ODE.