Question

Differential equations
Solve yp^2+2xp-y=0​

Answer 1

Given,

The given equation is $\[y{p^2} + 2xp - y = 0\].$

Solution,

Solve the given equation.

$\[\begin{array}{l} \Rightarrow y{p^2} - 2xp + y = 0\\ \Rightarrow y = \frac{{2xp}}{{1 + {p^2}}}\end{array}\]$

Differentiate the above equation.

$\[ \Rightarrow \frac{{dy}}{{dx}} = p\]$

$\[ \Rightarrow p = \frac{{\left( {1 + {p^2}} \right)\left\{ {2p + 2x\frac{{dp}}{{dx}}} \right\} - 2xp\left( {2p} \right)\frac{{dp}}{{dx}}}}{{{{\left( {1 + {p^2}} \right)}^2}}}\]$

$\[ \Rightarrow p{\left( {1 + {p^2}} \right)^2} = 2p\left( {1 + {p^2}} \right) + 2x\frac{{dp}}{{dx}}\left( {1 + {p^2}} \right) - 4x{p^2}\frac{{dp}}{{dx}}\]$

$\[ \Rightarrow p{\left( {1 + {p^2}} \right)^2} = 2p\left( {1 + {p^2}} \right) + \frac{{dp}}{{dx}}2x\left( {1 + {p^2} - 2{p^2}} \right)\]$

Solve further,

$\[ \Rightarrow p{\left( {1 + {p^2}} \right)^2} = 2p\left( {1 + {p^2}} \right) + \frac{{dp}}{{dx}}\left( {1 - {p^2}} \right)2x\]$

$\[ \Rightarrow p{\left( {1 + {p^2}} \right)^2}\left( {1 + {p^2} - 2} \right) = \frac{{dp}}{{dx}}\left( {1 - {p^2}} \right)2x\]$

$\[ \Rightarrow p{\left( {1 + {p^2}} \right)^2}\left( {{p^2} - 1} \right) = \frac{{dp}}{{dx}}\left( {1 - {p^2}} \right)2x\]$

Neglect the negative terms.

$\[ \Rightarrow p\left( {1 + {p^2}} \right) + 2x\frac{{dp}}{{dx}} = 0\]$

Integrate the above.

$\[\begin{array}{l} \Rightarrow \int {\frac{{dx}}{x} + \int {\frac{{2dp}}{{p\left( {1 + {p^2}} \right)}} = 0} } \\ \Rightarrow x = \frac{{c\left( {1 + {p^2}} \right)}}{{{p^2}}}\end{array}\]$

Hence, the value is $\[\frac{{c\left( {1 + {p^2}} \right)}}{{{p^2}}}\].$

Answer 2

Answer:2cx+c^2=y^2

Step-by-step explanation:

y=2px+p^2y

2px=y-p^2y

   x =y-p^2y/2p

  x  =y/2p-p^2y/2p

   x =1/2(y/p)-1/2(py)

differtiate with respect to x

 1/p=1/2[-y/p^2dp/dy+1/2p] --[p/2+y/2 dp/dy]

1/2p+p/2= -dp/dy[y/2p^2+y/2]

=>dp/dy=-p/y

=>dp/p= -∫dy/y

=>logp= -logy+logc

=>p=c/y

putting the value of p in the question ,we get

y=2(c/y)x+c^2/y^2

y=2cx/y+c^2/y

y=2cx+c^2/y

y^2=2cx+c^2