If y = 3 cos (log x)+4 sin (log x), show that $x^{2} y_{2} + x y_{1} + y = 0.$

Open in App

Solution

Given, y = 3 cos (log x)+4 sin (log x) ....(i)

Differentiating w.r.t. x, we get

$\frac{d y}{d x} = y_{1} = - 3 s i n (l o g x) \frac{d}{d x} (l o g x) + 4 c o s (l o g x) \frac{d}{d x} l o g x = - 3 s i n (l o g x) \frac{1}{x} + 4 c o s (l o g x) \frac{1}{x} (∵ \frac{d y}{d x} = y_{1}) M u l t i p l y i n g b y x, w e g e t x y_{1} = - 3 s i n (l o g x) + 4 c o s (l o g x) . . . (i i) A g a i n d i f f e r e n t i a t i n g w . r . t . x, w e o b t a i n x y_{2} + y_{1} .1 = - 3 c o s (l o g x) \frac{d}{d x} (l o g x) - 4 s i n (l o g x) \frac{d}{d x} l o g x = - 3 c o s (l o g x) \frac{1}{x} - 4 s i n (l o g x) \frac{1}{x} M u l t i p l y i n g t h r o u g h o u t b y x, w e h a v e x^{2} y_{2} + x y_{1} = - 3 (3 c o s (l o g x) + 4 s i n (l o g x)) [f r o m E q . (i)] \Rightarrow x^{2} y_{2} + x y_{1} = - y \Rightarrow x^{2} y_{2} + x y_{1} + y = 0$

Suggest Corrections

Similar questions

If y = 3 cos (log x) + 4 sin (log x), show that $x^{2} y_{2} + x y_{1} + y = 0$
where $y_{1}$ & $y_{2}$ are the successive derivatives of y.

y = 3 cos (log x) + 4 sin (log x)

x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + y = 0

x = \sin (\frac{1}{a} \log y)

y = e^{a {cos}^{- 1} x}, - 1 \leq x \leq 1,

(1 - x^{2}) y_{2} - x y_{1} - a^{2} y = 0

Join BYJU'S Learning Program