Question

Solve :

x ²p + y ²q = z² .​

Answer 1

Answer:

First thought: Maybe we can transform it somehow into one of those f(p,q)=0f(p,q)=0 form by making some substitution. Let’s think.

To begin with, we can rewrite it this way…

(xz∂z∂x)2+(yz∂z∂y)2=1(1)(1)(xz∂z∂x)2+(yz∂z∂y)2=1

We seem to be stuck. It’s time to improvise! Let’s make some substitutions…

X=logxX=log⁡x

Y=logyY=log⁡y

Z=logzZ=log⁡z

Using the above substitutions, we have these relations…

∂Z∂X=xz∂z∂x∂Z∂X=xz∂z∂x

∂Z∂Y=yz∂z∂y∂Z∂Y=yz∂z∂y

Let’s rewrite equation (1)(1) in terms of X,YX,Y and ZZ

(∂Z∂X)2+(∂Z∂Y)2=1(∂Z∂X)2+(∂Z∂Y)2=1

Now we’ll let p=∂Z∂Xp=∂Z∂X and q=∂Z∂Yq=∂Z∂Y

So we have pretty little equation now…

p2+q2=1(2)(2)p2+q2=1

Let’s say the required solution of the above equation is…

Z=aX+bY+cZ=aX+bY+c

That means on partially differentiating with respect to XX and YY , we’ll have ∂Z∂X=a=p∂Z∂X=a=p and ∂Z∂Y=b=q∂Z∂Y=b=q

So from equation (2)(2) , we have…

a2+b2=1⟹b=1−a2−−−−−√a2+b2=1⟹b=1−a2

Using this value of bb , our proposed solution becomes…

Z=aX+1−a2−−−−−√Y+cZ=aX+1−a2Y+c

Finally, we undo our substitutions to get the final solution for our Partial Differential Equation.

logz=alogx+1−a2−−−−−√logy+c

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