Question

solve p square +q square =x+y​

Answer 1

p square plus q square is equal to x plus y

Answer 2

Therefore the required solution for the given Partial Differential Equation p² + q² = x + y is " 3z = ± 2 $( x + a ) ^3^/^2$ ± 2 $(y-a)^3^/^2$ + 3c ".

Given:

The partial differential equation: p² + q² = x + y

To Find:

The solution for the partial differential equation p² + q² = x + y

Solution:

The given question can be solved as shown below.

The given partial differential equation is p² + q² = x + y ____Equation-A

It is a Non-Linear First order Partial Differential Equation of the form f₁ ( x,p ) = f₂ ( y,q )

Now after re-writing the equation we get p² - x =y - q²

Let p² - x =y - q² = a

Case-1:

⇒ p² - x = a

⇒ p² = ( x + a )

⇒ p = ± √( x + a )

Case-2:

⇒ y - q² = a

⇒ q² = y - a

⇒ q = ± √ ( y - a )

Now, as we know that the complete solution of equation-A can be written as dz = pdx + qdy _____Equation B

Now put the value of p + q in equation-2, we get,

⇒ dz = ± √( x + a ) dx ± √ ( y - a ) dy

Integrating both sides we get,

⇒ ∫ dz = ± ∫ √( x + a ) dx ± ∫ √ ( y - a ) dy

⇒ z = ± $( x + a ) ^3^/^2 /( 3/2)$ ± $(y-a)^3^/^2/(3/2)$ + c

⇒ z = ± ( 2/3 ) $( x + a ) ^3^/^2$ ± ( 2/3) $(y-a)^3^/^2$ + c

⇒ 3z = ± 2 $( x + a ) ^3^/^2$ ± 2 $(y-a)^3^/^2$ + 3c

where a and b are Arbitrary constants

Therefore the required solution for the given Partial Differential Equation p² + q² = x + y is " 3z = ± 2 $( x + a ) ^3^/^2$ ± 2 $(y-a)^3^/^2$ + 3c ".

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