Answer to Question #152177 in Differential Equations for Ashweta Padhan

Question

Find a solution of p+q^2=0 passing through the line x0(s) =0, y0(s)=s, z0(s)=3s

Answer 1

Let z=f(x+cy) be the trial solution of the given PDE

Let us take u= x+cy, so z=f(u)

Then we have p=dz/du, q=cdz/du

q/p=c

q=pc

p+p²c²=0 p(1+pc²)=0

1+pc²=0

p=-1/c²

dz=pdx+qdy=p(dx+cdy)=0

dx+cdy=0

dx=-cdy

x=-cy+c₁

Substituting the value of p

p, we get

$dz=\frac{-(dx+cdy)}{c^2}$ =0

z=-x/c²-y/c+c₁

3s=-s/c+c₁

s/c=c₁-3s

c=s(c₁-3s)

c₁=3s+s/c

0=-cs+3s+s/c

0=-c²s+3cs+s

D=9s² +4s

$c=\frac{-3s+\sqrt{9s^2+4s}}{-2}$

Answer: $\frac{4}{(-3s+\sqrt{9s^2+4s})^2}-(\frac{2}{(-3s+\sqrt{9s^2+4s})})^2=0$