Question
Show that $\text{Ax}^2+\text{By}^2=1$ is a solution of the differential equation $\text{x}\Big\{\text{y}=\text{x}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dx}}{\text{dy}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}.$
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Answer
We have,
$\text{Ax}^2+\text{By}^2=1\ ...(1)$
Differentiating it with respect in x
$2\text{Ax}+2\text{By}\frac{\text{dy}}{\text{dx}}=0$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{-2\text{Ax}}{2\text{B}}$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{Ax}}{\text{B}}$
Differentiating it with respect in x
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=-\frac{\text{A}}{\text{B}}$
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}$
Using equation (1)
$\text{x}\Big\{\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}$
$\text{Ax}^2+\text{By}^2=1\ ...(1)$
Differentiating it with respect in x
$2\text{Ax}+2\text{By}\frac{\text{dy}}{\text{dx}}=0$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{-2\text{Ax}}{2\text{B}}$
$\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{Ax}}{\text{B}}$
Differentiating it with respect in x
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=-\frac{\text{A}}{\text{B}}$
$\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2=\frac{\text{y}}{\text{x}}\frac{\text{dy}}{\text{dx}}$
Using equation (1)
$\text{x}\Big\{\text{y}\frac{\text{d}^2\text{y}}{\text{dx}^2}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big\}=\text{y}\frac{\text{dy}}{\text{dx}}$
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