\(cos^{-1}(\frac{y}{2})=log_e(\frac{x}{5})^5, |y| < 2\)
Differentiate on both side
\(-\frac{ 1}{\sqrt{1-(\frac{y}{2})^2}} \times \frac{y'}{2} =\frac{ 5}{\frac{x}{5}} \times \frac{1}{5}\)
\(-\frac{xy'}{2} = 5 \sqrt{1-(\frac{y}{2})^2}\)
Square on both side
\(\frac{x^2y'^2}{4} = 25\bigg(\frac{4-y^2}{4}\bigg)\)
Diff on both side
\(2xy'^2+2y'y''x^2=-25 \times 2yy'\)
\(xy'+y''x^2+25y = 0\)
Hence, the correct option is (D): \(x^2y′′ + xy′+ 25y = 0\)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
Various trigonometric identities are as follows:
Cosecant and Secant are even functions, all the others are odd.
T-Ratios of (2x)
sin2x = 2sin x cos x
cos 2x = cos2x – sin2x
= 2cos2x – 1
= 1 – 2sin2x
T-Ratios of (3x)
sin 3x = 3sinx – 4sin3x
cos 3x = 4cos3x – 3cosx