If $l$ and $m$ are the degree and the order respectively of the differential equation of the family of all…

1. 5
2. 10
3. 15
4. 7

Solution

Family of all circles in the $x y$-planewith radius 5 units with center $\left(x_1, y_1\right)$ is $ \left(x-x_1\right)^2+\left(y-y_1\right)^2=5^2 $

$ \Rightarrow \quad\left(x-x_1\right)=-\left(y-y_1\right) y^{\prime} $ Now Eq. (ii) diff again w.r.t ' $x$ ', we get $ \begin{aligned} 2+2\left(y^{\prime \prime}\left(y-y_1\right)+\left(y^{\prime}\right)^2\right] & =0 \\ \Rightarrow \quad\left(y-y_1\right) & =\frac{1-\left(y^{\prime}\right)^2}{y^{\prime \prime}} \end{aligned} $ Sub. values in Eq. (i) $ \begin{aligned} =\left[\frac{1+\left(y^{\prime}\right)^2}{y^{\prime \prime}} \cdot y^{\prime}\right\rceil^2+\left[\frac{1+\left(y^{\prime}\right)^2}{y^{\prime \prime}}\right] & =25 \\ \Rightarrow \quad \frac{\left(1+y^{\prime 2}\right)}{\left(y^{\prime \prime}\right)^2} \cdot\left(y^{\prime}\right)^2+\frac{1+\left(y^{\prime}\right)^2}{\left(y^{\prime \prime}\right)^2} & =25 \\ \Rightarrow \quad \frac{\left(1+y^{\prime}\right)^3}{\left(y^{\prime \prime}\right)^2} & =25 \\ \Rightarrow 25\left(y^{\prime \prime}\right)^2=\left[1+\left(y^{\prime}\right)^2\right]^3 & \end{aligned} $ So, order $=2$ and degree $=2$ $ \begin{aligned} & \therefore \quad l=2 \text { and } m=2 \\ & \text { Now, } \quad 2 l+3 m=2 \times 2+3 \times 2 \\ & =4+6=10 \\ & \end{aligned} $

Asked in: AP EAMCET 2018 (23 Apr Shift 1)

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