Question

In each of the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:

(i) y = ex + 1	y'' − y' = 0
(ii) y = x2 + 2x + C	y' − 2x − 2 = 0
(iii) y = cos x + C	y' + sin x = 0
(iv) y = $\sqrt{1+x^2}$	y' = $\frac{xy}{1+x^2}$
(v) y = x sin x	xy' = y + x $\sqrt{x^2-y^2}$
(vi) $y=\sqrt{a^2-x^2}$	$x+y\frac{dy}{dx}=0$

Answer 1

(i) We have,
y'' − y' = 0 .....(1)
Now,
y = e^x +1

^{$$\begin{gathered} \Rightarrow y\text{'}=e^x \\ \Rightarrow y\text{'}\text{'}=e^x \end{gathered}$$}

Putting the above values in (1), we get
$\mathrm{LHS}=e^x-e^x=0=\mathrm{RHS}$
Thus, y = e^x +1 is the solution of the given differential equation.

(ii) We have,
y' − 2x − 2 = 0 .....(1)
Now,
y = x² + 2x + C
^{$\Rightarrow y\text{'}=2x+2$}
Putting the above value in (1), we get
$\mathrm{LHS}=2x+2-2x-2=0=\mathrm{RHS}$
Thus, y = x² + 2x + C is the solution of the given differential equation.

(iii) We have,
y' + sin x = 0 .....(1)
Now,
y = cos x + C
^{$\Rightarrow y\text{'}=-\sin x$}
Putting the above value in (1), we get
$\mathrm{LHS}=-\sin x+\sin x=0=\mathrm{RHS}$
Thus, y = cos x + C is the solution of the given differential equation.

(iv) We have,
y' = $\frac{xy}{1+x^2}$ .....(1)
Now,
y = $\sqrt{1+x^2}$
^{$\Rightarrow y\text{'}=\frac{x}{\sqrt{1+x^2}}$}

Putting the above value in (1), we get
$$\begin{gathered} \mathrm{LHS}=\frac{x}{\sqrt{1+x^2}} \\ =\frac{x}{\sqrt{1+x^2}}\times \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}} \\ =\frac{xy}{1+x^2}=\mathrm{RHS} \end{gathered}$$
Thus, y = $\sqrt{1+x^2}$ is the solution of the given differential equation.

(v) We have,
xy' = y + x $\sqrt{x^2-y^2}$ .....(1)
Now,
y = x sin x
^{$\Rightarrow y\text{'}=\sin x+x\cos x$}
Putting the above value in (1), we get
$$\begin{gathered} \mathrm{LHS}=x\left(\sin x+x\cos x\right) \\ =x\sin x+x^2\cos x \\ =x\sin x+x\left(x\cos x\right) \\ =x\sin x+x\left(x\sqrt{1-{\sin }^{2}x}\right) \\ =x\sin x+x\left(\sqrt{x^2-x^2{\sin }^{2}x}\right) \\ =y+x\left(\sqrt{x^2-y^2}\right)=\mathrm{RHS} \end{gathered}$$
Thus, y = x sin x is the solution of the given differential equation.


(v) We have,
xy' = y + x $\sqrt{x^2-y^2}$ .....(1)
Now,
y = x sin x
^{$\Rightarrow y\text{'}=\sin x+x\cos x$}
Putting the above value in (1), we get
$$\begin{gathered} \mathrm{LHS}=x\left(\sin x+x\cos x\right) \\ =x\sin x+x^2\cos x \\ =x\sin x+x\left(x\cos x\right) \\ =x\sin x+x\left(x\sqrt{1-{\sin }^{2}x}\right) \\ =x\sin x+x\left(\sqrt{x^2-x^2{\sin }^{2}x}\right) \\ =y+x\left(\sqrt{x^2-y^2}\right)=\mathrm{RHS} \end{gathered}$$
Thus, y = x sin x is the solution of the given differential equation.

(vi) We have,
$x+y\frac{dy}{dx}=0$ .....(1)

Now,
$y=\sqrt{a^2-x^2}$
^{$\Rightarrow y\text{'}=\frac{-x}{\sqrt{a^2-x^2}}$}

Putting the above value in (1), we get

$$\begin{gathered} \mathrm{LHS}=x+y\left(\frac{-x}{\sqrt{a^2-x^2}}\right) \\ =x+\sqrt{a^2-x^2}\left(\frac{-x}{\sqrt{a^2-x^2}}\right) \\ =x+\sqrt{a^2-x^2}\left(\frac{-x}{\sqrt{a^2-x^2}}\right) \\ =x-x=0=\mathrm{RHS} \end{gathered}$$

Thus, $y=\sqrt{a^2-x^2}$ is the solution of the given differential equation.