Question

Solve y = ptanp + log(cosp) where p = dy/dx

Answer 1

Given : y = ptanp + log(cosp) where p = dy/dx

To find : x in terms of p

Solution:

y = ptanp + log(cosp)

p = dy/dx

differentiating wrt x

=> dy/dx  = p sec²p (dp/dx)  + tanp (dp/dx)  +  (1/Cosp)(-Sinp)(dp/dx)

=> dy/dx  = p sec²p (dp/dx)  + tanp (dp/dx)  - tanp(dp/dx)

=> dy/dx  = p sec²p (dp/dx)

dy/dx = p

=> p = p sec²p (dp/dx)

=> 1 = sec²p (dp/dx)

=> dx = sec²p (dp)

integrating both sides

=> x  = tanp + c

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Answer 2

In parametric form, the solution is

$y=t\tan t+\log (\cos t)$ and $x=\tan t-c$,    (where t is a parameter)

In non-parametric form, the solution is

$y=(x+c)\tan^{-1}(x+c)+\log (\cos (\tan^{-1}(x+c)))$

Step-by-step explanation:

Given

$y=p\tan p+\log (\cos p)$

Differentiating the above w.r.t. $x$

$\frac{dy}{dx}=p\sec^2p\frac{dp}{dx}+\tan p\frac{dp}{dx}+\frac{1}{\cos p}\times(-\sin p)\frac{dp}{dx}$

$\implies p=p\sec^p\frac{dp}{dx}+\tan p\frac{dp}{dx}-\tan p\frac{dp}{dx}$

$\implies p=p\sec^2p\frac{dp}{dx}$

$\implies \sec^2p dp=dx$

$\implies\int \sec^2p dp=\int dx$

$\implies \tan p=x+c$    where c is a  constant

$\implies x=\tan p-c$

Thus,

$y=p\tan p+\log (\cos p)$ and $x=\tan p-c$ is parametric solution of the equation, where p is a parameter

If we eliminate p

$\tan p=x+c$

$\implies p=\tan^{-1}(x+c)$

Thus,

$y=\tan^{-1}(x+c)\tan (\tan^{-1}(x+c))+\log (\cos (\tan^{-1}(x+c)))$

$\implies y=(x+c)\tan^{-1}(x+c)+\log (\cos (\tan^{-1}(x+c)))$

Hope this answer is helpful.

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