Question

✪ Qᴜᴇsᴛɪᴏɴ:-


Vᴇʀɪғʏ ᴛʜᴀᴛ ᴛʜᴇ ғᴜɴᴄᴛɪᴏɴ ʏ = ᴀ ᴄᴏs x + ʙ sɪɴ x, ᴡʜᴇʀᴇ, ᴀ, ʙ ∈ R ɪs ᴀ sᴏʟᴜᴛɪᴏɴ ᴏғ ᴛʜᴇ ᴅɪғғᴇʀᴇɴᴛɪᴀʟ ᴇϙᴜᴀᴛɪᴏɴ ᴅ2ʏ/ᴅx2 + ʏ=0.​

Answer 1

Answer:

here is your answer buddy

Answer 2

✪ Qᴜᴇsᴛɪᴏɴ:-  

Vᴇʀɪғʏ ᴛʜᴀᴛ ᴛʜᴇ ғᴜɴᴄᴛɪᴏɴ ʏ = ᴀ ᴄᴏs x + ʙ sɪɴ x, ᴡʜᴇʀᴇ, ᴀ, ʙ ∈ R ɪs ᴀ sᴏʟᴜᴛɪᴏɴ ᴏғ ᴛʜᴇ ᴅɪғғᴇʀᴇɴᴛɪᴀʟ ᴇϙᴜᴀᴛɪᴏɴ ᴅ2ʏ/ᴅx2 + ʏ=0.​

✪ ANSWER:-

Given , y = A cos x + B sin x where A , B ∈ R

$\implies \bold{\frac{dy}{dx}=A(-sinx)+B(cosx) }\\\\\implies \bold{\frac{dy}{dx}=B.cosx-A.sinx }$

Now do d²y/dx².

$\implies \bold{\frac{d^2y}{dx^2}=B(-sinx)-A(cosx) }\\\\\implies \bold{\frac{d^2y}{dx^2}=-B.sinx-A.cosx }$

Now our required d²y/dx²+y.

$\implies \bold{\frac{d^2y}{dx^2}+y=(-B.sinx-A.cosx)+(A.cosx+B.sinx) }\\\\\implies \bold{\frac{d^2y}{dx^2}=A.cosx-A.cosx+B.sinx-B.sinx}\\\\\implies \bold{\bf{\red{\frac{d^2y}{dx^2}=0}}}$

Hence verified