Question

For each of the following differential equations verify that the accompanying function is a solution:

Differential equation	Function
(i) $x\frac{dy}{dx}=y$	y = ax
(ii) $x+y\frac{dy}{dx}=0$	$y=\pm \sqrt{a^2-x^2}$
(iii) $x\frac{dy}{dx}+y=y^2$	$y=\frac{a}{x+a}$
(iv) $x^3\frac{d^{2}y}{d{x}^2}=1$	$y=ax+b+\frac{1}{2x}$
(v) $y={\left(\frac{dy}{dx}\right)}^2$	$y=\frac{1}{4}{\left(x\pm a\right)}^2$

Answer 1

(i) We have,
$y=ax.....\left(1\right)$
Given differential equation: $x\frac{dy}{dx}=y$
Differentiating both sides of (1) with respect to x, we get
$$\begin{gathered} \frac{dy}{dx}=a \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}\left[\mathrm{Using}\left(1\right)\right] \\ \Rightarrow x\frac{dy}{dx}=y \end{gathered}$$
Hence, the given function is the solution to the given differential equation.

(ii) We have,
$$\begin{gathered} y=\pm \sqrt{a^2-x^2} \\ \Rightarrow y^2=a^2-x^2.....\left(1\right) \end{gathered}$$
Given differential equation: $x+y\frac{dy}{dx}=0$
Differentiating both sides of (1) with respect to x, we get
$$\begin{gathered} 2y\frac{dy}{dx}=-2x \\ \Rightarrow y\frac{dy}{dx}=-x \\ \Rightarrow x+y\frac{dy}{dx}=0 \end{gathered}$$
Hence, the given function is the solution to the given differential equation.

(iii) We have,
$$\begin{gathered} y=\frac{a}{x+a} \\ \Rightarrow xy+ay=a \\ \Rightarrow xy=a\left(1-y\right) \\ \Rightarrow \frac{xy}{1-y}=a \\ \Rightarrow \frac{1-y}{xy}=\frac{1}{a}.....\left(1\right) \end{gathered}$$
given differential equation: $x\frac{dy}{dx}+y=y^2$
Differentiating both sides of (1) with respect to x, we get
$$\begin{gathered} \frac{xy\left(0-{\displaystyle \frac{dy}{dx}}\right)-\left(1-y\right)\left(x{\displaystyle \frac{dy}{dx}}+y\right)}{{\left(xy\right)}^2}=0 \\ \Rightarrow xy\left(-\frac{dy}{dx}\right)-\left(1-y\right)\left(x\frac{dy}{dx}+y\right)=0 \\ \Rightarrow -xy\frac{dy}{dx}-x\frac{dy}{dx}-y+xy\frac{dy}{dx}+y^2=0 \\ \Rightarrow -x\frac{dy}{dx}-y+y^2=0 \\ \Rightarrow x\frac{dy}{dx}+y=y^2 \end{gathered}$$
Hence, the given function is the solution to the given differential equation.

(iv) We have,
$y=ax+b+\frac{1}{2x}.....\left(1\right)$
Differentiating both sides of (1) with respect to x, we get
$$\begin{gathered} \frac{dy}{dx}=a-\frac{1}{2x^2}.....\left(2\right) \\ \mathrm{Now}\mathrm{differentiating}\mathrm{both}\mathrm{sides}\mathrm{of}\left(2\right)\mathrm{with}\mathrm{respect}\mathrm{to}x,\mathrm{we}\mathrm{get} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\left(-\frac{1}{2}\right)\times \left(\frac{-2}{x^3}\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{x^3} \\ \Rightarrow x^3\frac{d^{2}y}{d{x}^2}=1 \end{gathered}$$
Hence, the given function is the solution to the given differential equation.

(v) We have,
$y=\frac{1}{4}{\left(x\pm a\right)}^2.....\left(1\right)$
Differentiating both sides of (1) with respect to x, we get
$$\begin{gathered} \frac{dy}{dx}=\frac{1}{4}\times 2\left(x\pm a\right) \\ \Rightarrow \frac{dy}{dx}=\frac{1}{2}\left(x\pm a\right) \\ \mathrm{Squaring}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get} \\ \Rightarrow {\left(\frac{dy}{dx}\right)}^2={\left[\frac{1}{2}\left(x\pm a\right)\right]}^2 \\ \Rightarrow {\left(\frac{dy}{dx}\right)}^2=\frac{1}{4}{\left(x\pm a\right)}^2 \\ \Rightarrow {\left(\frac{dy}{dx}\right)}^2=y\left[\mathrm{Using}\left(1\right)\right] \\ \therefore y={\left(\frac{dy}{dx}\right)}^2 \end{gathered}$$
Hence, the given function is the solution to the given differential equation.