Question

solve y=2px+ysquare p square

Answer 1

SOLUTION

TO SOLVE

$\displaystyle \sf{y = 2px + {y}^{2} {p}^{2} }$

EVALUATION

Here the given differential equation is

$\displaystyle \sf{y = 2px + {y}^{2} {p}^{2} } \: \: \: - - - - (1)$

We solve it as below

$\displaystyle \sf{y = 2px + {y}^{2} {p}^{2} }$

$\displaystyle \sf{ \implies \: \: x = \frac{y}{2p} - \frac{p {y}^{2} }{2} }$

Differentiating both sides with respect to y we get

$\displaystyle \sf{ \implies \: \: \frac{dx}{dy} = \frac{1}{2p} - \frac{y}{ {p}^{2} } \frac{dp}{dy} - py - \frac{ {y}^{2} }{2} \frac{dp}{dy} }$

$\displaystyle \sf{ \implies \: \: \frac{1}{p} = \frac{1}{2p} - \frac{y}{ {p}^{2} } \frac{dp}{dy} - py - \frac{ {y}^{2} }{2} \frac{dp}{dy} }$

$\displaystyle \sf{ \implies \: \: \frac{1}{p} = - \frac{y}{ {p}^{2} } \frac{dp}{dy} - py - \frac{ {y}^{2} }{2} \frac{dp}{dy} }$

$\displaystyle \sf{ \implies \: \: \frac{1}{p} + py= - \frac{y}{ 2p } \frac{dp}{dy} \bigg(\frac{1}{p} + py \bigg) }$

$\displaystyle \sf{ \implies \: \: \bigg(\frac{1}{p} + py \bigg) + \frac{y}{ 2p } \frac{dp}{dy} \bigg(\frac{1}{p} + py \bigg) = 0 }$

$\displaystyle \sf{ \implies \: \: \bigg(\frac{1}{p} + py \bigg) \bigg(1 + \frac{y}{ 2p } \frac{dp}{dy} \bigg) = 0 }$

$\displaystyle \sf{ \bigg(\frac{1}{p} + py \bigg) = 0 \: \: leads \: to \: singular \: solution }$

$\displaystyle \sf{ \: \bigg(1 + \frac{y}{ 2p } \frac{dp}{dy} \bigg) = 0 \: \: gives }$

$\displaystyle \sf{ \implies \: \frac{dp}{p} = - 2 \frac{dy}{y} }$

On integration we get

$\displaystyle \sf{ \int \frac{dp}{p} = - 2 \int \frac{dy}{y} }$

$\displaystyle \sf{ \implies \log \: p= - 2 \log y + \log c}$

$\displaystyle \sf{ \implies \log \: p= \log {y}^{ - 2} + \log c}$

$\displaystyle \sf{ \implies p= \frac{c}{ {y}^{2} } }$

Putting the value of p in Equation 1 we get

$\displaystyle \sf{y = \frac{2cx}{ {y}^{2} } + {y}^{2} \frac{ {c}^{2} }{ {y}^{4} } }$

$\displaystyle \sf{ \implies \: y = \frac{2cx}{ {y}^{2} } + \frac{ {c}^{2} }{ {y}^{2} } }$

$\displaystyle \sf{ \implies \: {y}^{3} = 2cx + {c}^{2} }$

FINAL ANSWER

Hence the required solution is

$\boxed{ \: \: \displaystyle \sf{ \: {y}^{3} = 2cx + {c}^{2} } \: \: }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

1. Solve the differential equation :-

(cosx.cosy - cotx) dx - (sinx.siny) dy=0

https://brainly.in/question/23945760

2. Find the complete integral of p-q=0

https://brainly.in/question/23947735