Question

The partial differential equation p+r+s=1 is of degree

Answer 1

The given differential equation $p+r+s=1$ is of degree $1$

Answer 2

The partial differential equation p + r + s = 1 is of degree 2

Given :

The partial differential equation p + r + s = 1

To find :

The degree of the partial differential equation

Solution :

Step 1 of 2 :

Write down the given partial differential equation

Here the given partial differential equation is

p + r + s = 1

Step 2 of 2 :

Find degree of the partial differential equation

$\displaystyle \sf{ }p + r + s = 1$

$\displaystyle \sf{ \implies } \frac{ \partial z}{\partial x} + \frac{ {\partial }^{2}z }{\partial {x}^{2} } + \frac{ {\partial }^{2} z}{\partial x \partial y } = 1$

We know that order of a partial differential equation is the order of the highest derivative appearing in it.

We see that the highest order derivative in the given partial differential equation is 2

Hence the partial differential equation p + r + s = 1 is of degree 2

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