Question

Q1.Solve the equation

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \\ \bf {x}^{2} \dfrac{ {d}^{2}y}{ {dx}^{2} } + 3x \dfrac{dy}{dx} + y = \dfrac{1}{ {(1 - x)}^{2} } \\ \end{gathered}\end{gathered}\end{gathered} \end{gathered}$​

Answer 1

Step-by-step explanation:

792-1=791, 791-7=784, 784-19= 765, 765-37=728, 728-61=667, 667-91=576, 576-127=449, 449-169=280,80-217=63.

The next number to be subtracted is 331 which is greater than 63.

The remainder got is not zero.

∴ 792 is not a perfect cube.

Answer 2

Step-by-step explanation:

Here (u) distance of object from mirror= -10

Focal length= -(1/2 of radius of curvature)

= -(1/2* 12) =-6

So by mirror formula we get,

1/F=1/V+1/U

-1/6=1/V-1/10

1/V=1/10–1/6

1/V=3/30 -5/30

1/V= -2/30

V= -15

DISTANCE OF IMAGE FROM MIRROR IS 15 cm left side of mirror

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \\ \bf {x}^{2} \dfrac{ {d}^{2}y}{ {dx}^{2} } + 3x \dfrac{dy}{dx} + y = \dfrac{1}{ {(1 - x)}^{2} } \\ \end{gathered}\end{gathered}\end{gathered} \end{gathered}$