Question

Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: y-cos y=x :(ysin y+cosy+x)y'=y

Answer 1

y-cos y=x.….(1)
Differentiating both sides of the equation with respect to x,we get:
$\frac{dy}{dx}-\frac{d}{dx}$(cosy)=$\frac{d}{dx}(x)$
⇒y'+sin y.y'=1
⇒y'(1+siny)=1
$⇒y'=\frac{1}{1+siny}$
Substituting the value of y'in equation(1),we get:
L.H.S.=(ysiny+cosy+x)y'
=(ysiny+cosy+y-cosy)×$\frac{1}{1+siny}$
=y(1+siny).$\frac{1}{1+siny}$
=y
=R.H.S.
Hence,the given function is the solution of the corresponding differential equation.