Question

solve the following equation p=tan(x-p/1+p^2)​

Answer 1

Answer:

This exercise in (non linear ODE. of first order and higher degree - solve in x)

p=tan(x−p1+p2)

My try:

x=tan−1(p)+p1+p2

Derivative for y:

dxdy=(11+p2+1−p2(1+p2)2)dpdy

p=dydx

So

dxdy=1p

Then

1p=1−p2(1+p2)2dpdy

dy=p−p31+p2dp

∫dy=∫p1+p2dp−∫p31+p2

y=12ln(1+p2)−12p2+12ln(1+p2)

y=12p2+ln(1+p2)

so the general solution is

x=tan−1(p)+p1+p2

y=12p2+ln(1+p2)

True ?

Thanks.

Step-by-step explanation:

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Answer 2

Answer:

solution--

tan^-1(p) = x-p/(1+p^2)

x = tan^-1(p) + p/(1+p^2) -----(1)

differentiate with respect to 'y'

dx/dy = 1/(1+p^2)dp/dy +{(1-p^2)/(1+p^2)^2}dp/dy

since dx/dy = 1/p

therefore,

1/p =[  1/(1+p^2) +{(1-p^2)/(1+p^2)^2}   ]dp/dy

dy = p[   1/(1+p^2) +{(1-p^2)/(1+p^2)^2}   ]dp

dy = p[( 1+p^4+2p^2+1-p^4) / (1+p^2)^2 ] dp

dy =p [2(  1+ p^2) / (1+p^2)^2 ] dp

now ,

dy = {2p/1+p^2}dp

on integrating

we get

y = 1/2 log(1+p^2)+ log c -----(2)

from eq. (1) & (2) we get our required answer

Step-by-step explanation: