Question

Solve this plzz y=p tan p +log (cos p) where p =dy /dx

Answer 1

Answer:

log tan y/4 = c – 2 cos x/2 (D) none of these ... Solution of dy/dx = y/x + tan y /x is: ... Solution of 2y sinx dy/dx = 2sinx cosx –y2cosx, x=p/2, y=1 is given by.

Step-by-step explanation:

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Answer 2

In parametric form, the solution is

$y=p\tan p+\log (\cos p)$ and $x=\tan p-c$

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. $x$

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

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

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

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

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

$\implies \tan p=x+c$    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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