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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Form the differential equation corresponding to $(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2$ by eliminating a and b.
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Answer

The equation of the family of curves is

$(\text{x}-\text{a})^2+(\text{y}-\text{b})^2=\text{r}^2\ ...(1)$

where a and b is a parameter.

This equation contains only one arbitrary constant, so we shall get a differential equation of first order.

Differentiating equation (1) with respect to x, we get

$2(\text{x}-\text{a})+2(\text{y}-\text{b})\frac{\text{dy}}{\text{dx}}=0\ ...(2)$

Differentiating equation (1) with respect to x, we get

$2+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+2(\text{y}-\text{b})\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$

$\Rightarrow1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+(\text{y}-\text{b})\frac{\text{d}^2\text{y}}{\text{dx}^2}=0$

$\Rightarrow(\text{y}-\text{b})=\frac{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}$

From (2) and (3), we get

$(\text{x}-\text{a})-\frac{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}\frac{\text{dy}}{\text{dx}}=0\Rightarrow(\text{x}-\text{a})=\frac{\frac{\text{dy}}{\text{dx}}+\Big(\frac{\text{dy}}{\text{dx}}\Big)^3}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}\ ...(4)$

From (1) and (3), we get

$\frac{\Big[\frac{\text{dy}}{\text{dz}}+\Big(\frac{\text{dy}}{\text{dz}}\Big)^2\Big]^2}{\Big(\frac{\text{d}^2\text{y}^2}{\text{dz}^2}\Big)^2}+\frac{\Big[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Big]^2}{\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2}=\text{r}^2$

$\Rightarrow\frac{\Bigg[\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^4+\Big(\frac{\text{dy}}{\text{dx}}\Big)^6\Bigg]+\Bigg[1+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dy}}{\text{dx}}\Big)^4\Bigg]}{\Big(\frac{\text{d}^2\text{y}}{\text{dx}}\Big)^2}$

$\Rightarrow\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^4+\Big(\frac{\text{dy}}{\text{dx}}\Big)^6+1+2\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+\Big(\frac{\text{dy}}{\text{dx}}\Big)^4=\text{r}^2\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)$

$\Rightarrow1+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^2+3\Big(\frac{\text{dy}}{\text{dx}}\Big)^4+\Big(\frac{\text{dy}}{\text{dx}}\Big)^6=\text{r}^2\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)$

$\Rightarrow\Bigg[1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2\Bigg]^3=\text{r}^3\Big(\frac{\text{d}^2\text{y}}{\text{dx}^2}\Big)^2$

It is the required differential equation.

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