Question

the complete solution of partial differential equation
p(1+q)=qz​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

\Rightarrow \ln (k z-1)=x+k y+c

Step-by-step explanation:

Concept

Partial differential equations are made up of a function with several unknown variables and their partial derivatives.

Solve the Partial differential equation

p(1+q)=q z

Here $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$

Assuming$p=\frac{\partial z}{\partial x}=a$, where a is some constant. Then we get $q=\frac{a}{z-a}$ Then

$d z=a d x+\frac{a}{z-a} d y$

For equations of type $\phi(z, p, q)=0$ (i.e., without x, y explicitly), assume $u=x+k y$ and $z=f(u)=f(x+k y)$.

Now, $p=\frac{d z}{d u} \cdot 1, q=\frac{d z}{d u} \cdot k$.

Here,

$\frac{d z}{d u}\left(1+k \frac{d z}{d u}\right)=k z \frac{d z}{d u}$

$\Rightarrow \frac{k d z}{k z-1}=d u$

$\Rightarrow \ln (k z-1)=u+c$

$\Rightarrow \ln (k z-1)=x+k y+c$

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