Question

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: $y-\cos y=x:\left( y\sin y+\cos y+x \right)y'=y$.

Answer 1

Differentiate the function $y-\cos y=x$ both the sides with respect to x. Use the formula: - $\dfrac{d\cos x}{dx}=-\sin x$, to simplify the derivative. Now, substitute the value of x and y’, i.e., $\dfrac{dy}{dx}$, in the expression present on the L.H.S. of the differential equation. Simplify this side and check whether we are getting y or not. If we are getting y then $y-\cos y=x$ is a solution of the given differential equation otherwise not.

Complete step-by-step answer:
Here, we have been provided with the differential equation: $\left( y\sin y+\cos y+x \right)y'=y$, and we have been asked to check whether $y-\cos y=x$ is a solution of the given differential equation or not.
Now, if $y-\cos y=x$ is a solution of the given differential equation then it must satisfy the differential equation. Now, substituting the value of x from $x=y-\cos y$ in the L.H.S. of the given differential equation, we get,
$\Rightarrow L.H.S.=\left( y\sin y+\cos y+y-\cos y \right)y'$
Cancelling the like terms, we get,
$\Rightarrow L.H.S.=\left( y\sin y+y \right)y'$
$\Rightarrow L.H.S.=y\left( 1+\sin y \right)y'$ - (1)
Now, we need to substitute the value of y’, i.e., $\dfrac{dy}{dx}$, in the above expression. So, let us find its value.
$\because x=y-\cos y$
Differentiating both the sides with respect to the variable x, we get,
$\Rightarrow 1=\dfrac{dy}{dx}-\dfrac{d\cos y}{dx}$
Using chain rule of differentiation, we get,

& \Rightarrow 1=\dfrac{dy}{dx}-\dfrac{d\left[ \cos y \right]}{dy}\times \dfrac{dy}{dx} \\\ & \Rightarrow 1=y'-\dfrac{d\left[ \cos y \right]}{dy}\times y' \\\ \end{aligned}$$ We know that $$\dfrac{d\left[ \cos y \right]}{dy}=-\sin y$$, so we get, $$\begin{aligned} & \Rightarrow 1=y'+\sin y\times y' \\\ & \Rightarrow 1=y'\left( 1+\sin y \right) \\\ & \Rightarrow y'=\left( \dfrac{1}{1+\sin y} \right) \\\ \end{aligned}$$ So, substituting the value of y’ in the equation (1), we get, $$\Rightarrow L.H.S.=y\left( 1+\sin y \right)\times \left( \dfrac{1}{1+\sin y} \right)$$ Cancelling the like terms, we get, $$\begin{aligned} & \Rightarrow L.H.S.=y \\\ & \Rightarrow L.H.S.=R.H.S. \\\ \end{aligned}$$ Hence, we can conclude that $$y-\cos y=x$$ is a solution of the given differential equation. **Note:** One may note that there can be other methods also to solve the above question. What we can do is we will solve the given differential equation by some properties of the exact differential equation and ILATE rule. Here, the constant of indefinite integration ‘c’ will be introduced. Now, we will find if the value of ‘c’ can be 1 or not because the obtained solution will be $$cy-\cos y=x$$. The condition c = 1 will be possible if $$y\left( \dfrac{\pi }{2} \right)=\left( \dfrac{\pi }{2} \right)$$. But remember that here we are not asked to solve the equation but we need to check only so it will be better if you will use the approach we have used in the solution.