P^3 - 2xyp + 4y^2 = 0 solve the differential equation
Answers
Concept-
First, separate the values of x and y means variables of x on one side and variables of y on the other side, then differentiate both the sides w.r.t. y and find the value of y.
Given-
Equation is given as : p³ - 2xpy + 4y² = 0
Find-
Solve the differential equation p³ - 2xpy + 4y² = 0
Solution-
p³ - 2xpy + 4y² = 0.......(a)
⇒ 2xyp = p³ + 4y²
⇒ 2x = p²/ y + 4y/p
differentiating w.r.t. y gives,
2(dx/dy) = [(2p/y) (dp/dy)] - p²/y² + 4[1/p - y/p² (dy/dx)] → equation (1)
As, dy / dx = p ⇒ dx / dy = 1/p
equation (1) ⇒ 2/p = [(2p/y)(dp/dy)] - [p²/y²] + 4[(1/p) - (y/p²) (dy/dp)]
⇒ [(2p/y)(dp/dy)] - p²/y² + 2/p - [(4y/p²)(dp/dy)] = 0
⇒ [p-2y( dp/dy)] [2y²-p³]=0
2y²-p³ gives singular solutions.
so, we consider,
⇒ [p-2y(dp/dy)] = 0
⇒ p² = cy , c=constant
(a) ⇒ p(p² - 2xy) = -4y²
⇒ p(cy - 2xy) = -4y² (as p² = cy)
⇒ 2x-c=4y/p
⇒ (2x-c)²=16y²/p²
⇒ (2x-c)²=16y²/cy (as p²=cy)
⇒ 16y = c(2x-c)
⇒ y=c(2x-c)²/16
Therefore, the value of p³ - 2xyp + 4y² = 0 is y=c(2x-c)²/16.
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