Question

Find the Differential Equation whose solution is $a^{2}y-ax+8=0$.

• A. $8{y^2}_1+y+x{y}_1=0$
• B. $8{y^2}_1+y-x{y}_1=0$
• C. $8{y^2}_1-y-x{y}_1=0$
• D. $8{y^2}_1-y+x{y}_1=0$

Answer 1

The correct option is B

$8{y^2}_1+y-x{y}_1=0$

Explanation for the correct option:

Find the required differential equation.

The solution of the required differential equation is given as,

$a^{2}y-ax+8=0.....\left(1\right)$

differentiate equation $\left(1\right)$ with respect to$x$ we get,

$$\begin{gathered} a^2\frac{dy}{dx}-a\frac{d\left(x\right)}{dx}+8=0 \\ \Rightarrow \frac{dy}{dx}{a}^2-a^{}\times 1+0=0 \\ \Rightarrow \frac{dy}{dx}{a}^2-a=0 \\ \Rightarrow a^2{y}_1-a=0\left(\because \frac{dy}{dx}=y_1\right) \\ \Rightarrow a{(}^{}a{y}_1-1)=0 \end{gathered}$$

So, $a=0$ and $a=\frac{1}{y_1}$

$a=0$ is not possible as the value of the arbitrary constant can not be $0$.

Therefore, $a=\frac{1}{y_1}$

By putting the value of $a$ in equation $\left(1\right)$ we get,

$\Rightarrow \frac{y}{{y_1}^2}-\frac{x}{y_1}+8=0$

By multiplying ${y_1}^2$both sides,

$\Rightarrow y-x{y}_1+8{y_1}^2=0$

$\Rightarrow 8{y_1}^2+y-x{y}_1=0$

Hence, option $\left(B\right)$ is the correct answer.