Question

find the general solution of yzp+xzp=xy​

Answer 1

xzp+yzq=xy

This is Lagrange's equation

Pp+Qq=R

The auxiliary equation is,

dx/P=dy/Q=dz/R

⇒dx/xz=dy/yz=dz/xy...(1)

⇒dx/xz=dy/yz⇒dx/x=dy/y

⇒∫dx/x=∫dy/y

⇒logx=logy+loga

⇒logx/y=loga

∴x/y=a

Also (1) is equal to,

ydx+xdy−2zdz/xyz+xyz−2zxy=ydx+xdy−2zdz/0

d(xy)−2z dz=0

∴xy−z^2=b

General solution is ϕ(x/y,xy−z^2)=0

Answer 2

Answer:

Step-by-step explanation:

From the above question we need to find the general solution of,

yzp + xzp = xy​

To solve this equation, we have to factor out z from the left-hand side:

So,

          yzp + xzp = xy

          zp(y + x) = xy

By dividing both sides by z(y + x), we will get:

          p = xy / z(y + x)

So the general solution for this equation is:

          p = xy / z(y + x)

Note that this solution assumes that z(y + x) is not equal to zero, as division by zero is undefined. If z(y + x) = 0.

The equation reduces to 0 = 0, which is true for any value of p.

The he general solution of yzp + xzp = xy​ is z(y + x) = 0.

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