Question

solve the partial differential equation of root under P + root under Q equal to 2x​

Answer 1

Step-by-step explanation:

2x=√p+√q

so

p+q +√(4pq)=4x^2

Answer 2

Answer:

The solution for partial differential equation $\sqrt{p}+\sqrt{q}=2x$ is $z=\frac{(\alpha +2x)^{3}}{3}+y\alpha^{2}+C$.

Step-by-step explanation:

Consider the partial differential equation as follows:

$\sqrt{p}+\sqrt{q}=2x$

⇒ $\sqrt{p}-2x=-\sqrt{q}$

Let $\alpha$ be any arbitrary such that $\sqrt{p}-2x=-\sqrt{q}=\alpha$.

This is of the form $F_{1}(x,p)=F_{2}(x,q)=\alpha$.

⇒ $\sqrt{p}-2x=\alpha$

⇒ $\sqrt{p}=\alpha+2x$

⇒ $p=(\alpha +2x)^{2}$ (Squaring both side) ...... (1)

$-\sqrt{q}=\alpha$

⇒ $q=a^{2}$ (Squaring both side) ...... (2)

Since $dz=pdx+qdy$

$dz=(\alpha +2x)^{2}dx+{\alpha}^{2}dy$ (From (1) and (2))

Integrate both the sides.

$z=\frac{(\alpha +2x)^{3}}{3}+y\alpha^{2}+C$, where $C$ is the integration constant.

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