Question

y^2logy=xpylogx+x^2p^2 by using logy=v,logx=u find G.S,S.S

Answer 1

Given equation is $y^2logy=xpylogx+x^2p^2$

Putting, $log y = v, logx = u$

then,

$\frac{1}{y}dy = dv, \frac{1}{x}dx = du$ $\implies \frac{dy}{dx} = \frac{y}{x}\frac{dv}{du}$

Putting the values, then equation will be,

$y^2(\frac{dv}{du})^2+xyu\frac{y}{x}\frac{dv}{du} - y^2v = 0$

$(\frac{dv}{du})^2+u\frac{dv}{du} - v = 0$

Let, $P = \frac{dv}{du}$ then $P^2 +uP-v = 0$

Differentiating equation, we get,

$2P\frac{dP}{du}+P+u\frac{dP}{du}-P = 0 \implies \frac{dP}{du}(2P+u) = 0$

Then, $\frac{dP}{du} = 0 \implies P = k$

$\frac{dv}{du} = k \implies v = ku$ (1)

Also, $2P+u = 0 \implies 2dv = -udu$

$2v = -\frac{1}{2}u^2+C$ (2)

From (1), $e^y = ke^x \implies y =Kx$ (3)

From (2), $2e^y = -\frac{1}{2}e^{2x} + C$ (4)

Equation(3) and (4) are the required solutions.