Solution

$y=e^{x}+1$

Differentiating both sides of this equation with respect to $x$, we get:

$\frac{d y}{d x}=\frac{d}{d x}(e^{x}+1)$

$\Rightarrow y^{\prime}=e^{x}$

Now, differentiating equation (1) with respect to $x$, we get:

$\frac{d}{d x}(y^{\prime})=\frac{d}{d x}(e^{x})$

$\Rightarrow y^{\prime \prime}=e^{x}$

Substituting the values of $y^{\prime}$ and $y^{\prime \prime}$ in the given differential equation, we get the L.H.S. as:

$y^{\prime \prime}-y^{\prime}=e^{x}-e^{x}=0=$ R.H.S.

Thus, the given function is the solution of the corresponding differential equation.

Solution

$y=x^{2}+2 x+C$

Differentiating both sides of this equation with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(x^{2}+2 x+C)$

$\Rightarrow y^{\prime}=2 x+2$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=y^{\prime}-2 x-2=2 x+2-2 x-2=0=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y=\cos x+C$

Differentiating both sides of this equation with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(\cos x+C)$

$\Rightarrow y^{\prime}=-\sin x$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=y^{\prime}+\sin x=-\sin x+\sin x=0=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y=\sqrt{1+x^{2}}$

Differentiating both sides of the equation with respect to $x$, we get: $y^{\prime}=\frac{d}{d x}(\sqrt{1+x^{2}})$

$y^{\prime}=\frac{1}{2 \sqrt{1+x^{2}}} \cdot \frac{d}{d x}(1+x^{2})$

$y^{\prime}=\frac{2 x}{2 \sqrt{1+x^{2}}}$

$y^{\prime}=\frac{x}{\sqrt{1+x^{2}}}$

$\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \times \sqrt{1+x^{2}}$

$\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \cdot y$

$\Rightarrow y^{\prime}=\frac{x y}{1+x^{2}}$

$\therefore$ L.H.S. $=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y=A x$

Differentiating both sides with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(A x)$

$\Rightarrow y^{\prime}=A$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=x y^{\prime}=x \cdot A=A x=y=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y=x \sin x$

Differentiating both sides of this equation with respect to $x$, we get:

$y^{\prime}=\frac{d}{d x}(x \sin x)$

$\Rightarrow y^{\prime}=\sin x \cdot \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\sin x)$

$\Rightarrow y^{\prime}=\sin x+x \cos x$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

$ \begin{aligned} \text{ L.H.S. }=x y^{\prime} & =x(\sin x+x \cos x) \\ & =x \sin x+x^{2} \cos x \\ & =y+x^{2} \cdot \sqrt{1-\sin ^{2}} x \\ & =y+x^{2} \sqrt{1-(\frac{y}{x})^{2}} \\ & =y+x \sqrt{y^{2}-x^{2}} \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

Solution

$x y=\log y+C$

Differentiating both sides of this equation with respect to $x$, we get: $\frac{d}{d x}(x y)=\frac{d}{d x}(\log y)$

$\Rightarrow y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}=\frac{1}{y} \frac{d y}{d x}$

$\Rightarrow y+x y^{\prime}=\frac{1}{y} y^{\prime}$

$\Rightarrow y^{2}+x y y^{\prime}=y^{\prime}$

$\Rightarrow(x y-1) y^{\prime}=-y^{2}$

$\Rightarrow y^{\prime}=\frac{y^{2}}{1-x y}$

$\therefore$ L.H.S. $=$ R.H.S.

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y-\cos y=x$

Differentiating both sides of the equation with respect to $x$, we get:

$ \begin{aligned} & \frac{d y}{d x}-\frac{d}{d x}(\cos y)=\frac{d}{d x}(x) \\ & \Rightarrow y^{\prime}+\sin y \cdot y^{\prime}=1 \\ & \Rightarrow y^{\prime}(1+\sin y)=1 \\ & \Rightarrow y^{\prime}=\frac{1}{1+\sin y} \end{aligned} $

Substituting the value of $y^{\prime}$ in equation (1), we get:

L.H.S. $=(y \sin y+\cos y+x) y^{\prime}$

$ \begin{aligned} & =(y \sin y+\cos y+y-\cos y) \times \frac{1}{1+\sin y} \\ & =y(1+\sin y) \cdot \frac{1}{1+\sin y} \\ & =y \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

Solution

$x+y=\tan ^{-1} y$

Differentiating both sides of this equation with respect to $x$, we get:

$ \frac{d}{d x}(x+y)=\frac{d}{d x}(\tan ^{-1} y) $

$ \Rightarrow 1+y^{\prime}=[\frac{1}{1+y^{2}}] y^{\prime} $

$\Rightarrow y^{\prime}[\frac{1}{1+y^{2}}-1]=1$

$\Rightarrow y^{\prime}[\frac{1-(1+y^{2})}{1+y^{2}}]=1$

$\Rightarrow y^{\prime}[\frac{-y^{2}}{1+y^{2}}]=1$

$\Rightarrow y^{\prime}=\frac{-(1+y^{2})}{y^{2}}$

Substituting the value of $y^{\prime}$ in the given differential equation, we get:

L.H.S. $=y^{2} y^{\prime}+y^{2}+1=y^{2}[\frac{-(1+y^{2})}{y^{2}}]+y^{2}+1$

$ \begin{aligned} & =-1-y^{2}+y^{2}+1 \\ & =0 \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

Solution

$y=\sqrt{a^{2}-x^{2}}$

Differentiating both sides of this equation with respect to $x$, we get:

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}(\sqrt{a^{2}-x^{2}}) \\ & \Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \cdot \frac{d}{d x}(a^{2}-x^{2}) \\ & =\frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x) \\ & =\frac{-x}{\sqrt{a^{2}-x^{2}}} \end{aligned} $

Substituting the value of $\frac{d y}{d x}$ in the given differential equation, we get:

L.H.S. $=x+y \frac{d y}{d x}=x+\sqrt{a^{2}-x^{2}} \times \frac{-x}{\sqrt{a^{2}-x^{2}}}$

$ \begin{aligned} & =x-x \\ & =0 \\ & =\text{ R.H.S. } \end{aligned} $

Hence, the given function is the solution of the corresponding differential equation.

Solution

We know that the number of constants in the general solution of a differential equation of order $n$ is equal to its order.

Therefore, the number of constants in the general equation of fourth order differential equation is four.

Hence, the correct answer is D.

Solution

In a particular solution of a differential equation, there are no arbitrary constants.

Hence, the correct answer is D.