Question

solve root p+root q=1​

Answer 1

Step-by-step explanation:

root 0 + root 1 is equal to 1

because root 0 is equal to 0 and root 1 is equal to 1

Answer 2

Answer:

The solution for partial differential equation $\sqrt{p}+\sqrt{q}=1$ is $z=x\alpha^2+y(1-\alpha)^2+C$.

Step-by-step explanation:

Consider the partial differential equation as follows:

$\sqrt{p}+\sqrt{q}=1$

⇒ $\sqrt{p}=1-\sqrt{q}$

Let $\alpha$ be any arbitrary such that $\sqrt{p}=1-\sqrt{q} = \alpha$.

This is the form of $F_1(x,p)=F_2(x,q)=\alpha$.

⇒ $\sqrt{p}=\alpha$

Squaring both sides, we get

⇒ $p=\alpha^2$ ...... (1)

Also, $1-\sqrt{q}=\alpha$

⇒ $\sqrt{q}=1-\alpha$

Squaring both sides, we get

⇒ $q=(1-\alpha)^2$ ...... (2)

Since $dz=pdx+qdy$

$dz=\alpha^2dx+(1-\alpha)^2dy$ (From (1) and (2))

Now, integrate both the sides.

$z=x\alpha^2+y(1-\alpha)^2+C$, where $C$ is the integration constant.

Therefore, the solution for partial differential equation $\sqrt{p}+\sqrt{q}=1$ is $z=x\alpha^2+y(1-\alpha)^2+C$.

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