MCQ
Solution of $\frac{{dy}}{{dx}} = \frac{{x\log {x^2} + x}}{{\sin y + y\,\,\cos y}}$ is
  • $y\sin y = {x^2}\log x + c$
  • B
    $y\sin y = {x^2} + c$
  • C
    $y\sin y = {x^2} + \log x + c$
  • D
    $y\sin y = x\log x + c$

Answer

Correct option: A.
$y\sin y = {x^2}\log x + c$
a
(a) $\frac{{dy}}{{dx}} = \frac{{x\log {x^2} + x}}{{\sin y + y\cos y}}$.

Separating the variables and integrating

$\int {(\sin y + y\cos y)dy = \int {(x\log {x^2} + x)dx} } $

==> $ - \cos y + y\sin y + \cos y$

$ = \frac{{{x^2}}}{2}\log {x^2} - \int {\frac{{{x^2}}}{2}.\frac{1}{{{x^2}}}.2xdx + \int {x\,dx + c} } $

==> $y\sin y = \frac{{{x^2}}}{2}2\log x - \int {x\,dx + \int {xdx + c} } $

==> $y\sin y = {x^2}\log x + c$.

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