Question

For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

y=cosy=x and (ysiny+cosy+x)y'=y.

Answer 1

Given, y-cosy=x
On differentiating both sides w.r.t. x, we get
$y^{\prime }-\frac{d}{dx}\left(cosy\right)=\frac{d}{dx}x\Rightarrow y^{\prime }-(-siny)y^{\prime }=1(\because \frac{dy}{dx}=y^{\prime })\Rightarrow y^{\prime }(1+siny)=1\Rightarrow y^{\prime }=\frac{1}{1+siny}$
But we have to verify, $(ysiny+cosy+x)y^{\prime }=y....\left(i\right)$
On substituting the value of y'in Eq. (i), we get
$LHS=(ysiny+cosy+x)y^{\prime }=(ysiny+y)\times \frac{1}{1+siny}(\because y=x+cosy)=y(1+siny)\times \frac{1}{1+siny}=y=RHS$
Hence, y-cosy=x is a solution of of the given differential equation.