solve p(1+q^2)=q(z-a) using charpit's method
Answers
Answer:
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Step-by-step explanation:
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Solution
Here, given equation is:
p(1+q^2)=q(z-a)\\ \therefore p+pq^2)=qz-qa\\ \implies p+pq^2-qz+qa=0---(1)p(1+q
2
)=q(z−a)
∴p+pq
2
)=qz−qa
⟹p+pq
2
−qz+qa=0−−−(1)
Therefore the Charpit's auxiliary equation is
\frac{dp}{\frac{df}{dx}+p\frac{df}{dz}}=\frac{dq}{\frac{df}{dy}+q\frac{df}{dz}}=\frac{dz}{-p\frac{df}{dp}-q\frac{df}{dq}}=\frac{dx}{-\frac{df}{dp}}=\frac{dy}{-\frac{df}{dq}}=\frac{dF}{0}
dx
df
+p
dz
df
dp
=
dy
df
+q
dz
df
dq
=
−p
dp
df
−q
dq
df
dz
=
−
dp
df
dx
=
−
dq
df
dy
=
0
dF
or
\frac{dp}{0+p(-q)}=\frac{dq}{0+q(-q)}=\frac{dz}{-p-pq^2}=\frac{dx}{-1-q^2}=\frac{dy}{-2pq-z}=\frac{dF}{0}
0+p(−q)
dp
=
0+q(−q)
dq
=
−p−pq
2
dz
=
−1−q
2
dx
=
−2pq−z
dy
=
0
dF
or
\frac{dp}{-pq}=\frac{dq}{-q^2}=\frac{dz}{-p-pq^2}=\frac{dx}{-1-q^2}=\frac{dy}{-2pq-z}=\frac{dF}{0}
−pq
dp
=
−q
2
dq
=
−p−pq
2
dz
=
−1−q
2
dx
=
−2pq−z
dy
=
0
dF
So that
\implies \frac{dp-dq}{-pq+q^2}=\frac{dy-dx}{-2pq-z+1+q^2}\\ \implies \frac{dp-dq}{-pq+q^2}=\frac{dy-dx}{-pq+q^2}\\ \implies dp-dq = dy-dx\\ \therefore dp+dx=dy+dq\\ ⟹
−pq+q
2
dp−dq
=
−2pq−z+1+q
2
dy−dx
⟹
−pq+q
2
dp−dq
=
−pq+q
2
dy−dx
⟹dp−dq=dy−dx
∴dp+dx=dy+dq
Integrating, we get
\implies p+x=y+q\\ \implies p=y-x+q\\⟹p+x=y+q
⟹p=y−x+q
From equation (1), we get
p+pq^2-qz+qa=0 \implies p=pq^2-qz+qa\\ q(pq-z-a)=p\\ q=p\ or\\ q=\frac{z-a}{p}p+pq
2
−qz+qa=0⟹p=pq
2
−qz+qa
q(pq−z−a)=p
q=p or
q=
p
z−a
Using the value pp and qq in dz=pdx+qdydz=pdx+qdy
\implies dz=(y-x+q)dx+(\frac{z-a}{p})dy\\ \implies \frac{dz}{z}=(y-x+q)dx+(\frac{z-a}{p})dy⟹dz=(y−x+q)dx+(
p
z−a
)dy
⟹
z
dz
=(y−x+q)dx+(
p
z−a
)dy
Integrating, we get
log\ z=yx+qx-\frac{x^2}{2}+\frac{y-ya}{py}log z=yx+qx−
2
x
2
+
py
y−ya