From the question it is given that,
y = A cos 2x – B sin 2x … [equation (i)]
Now, differentiate the equation (i) with respect x,
\( \dfrac{ dy}{dx}\)= – 2A sin (2x) – 2B cos 2x
Taking common terms outside,
\( \dfrac{ dy}{dx}\)= -2 (A sin 2x + B cos 2x) … [equation (ii)]
Then, the above equation is again differentiating with respect to x we get,
\( \dfrac{ d^2y}{dx^2}\)= – 2 [2A cos 2x – 2B sin 2x]
= -4 [A cos 2x – B sin 2x] … [equation (iii)]
The given differential equation is \( \dfrac{ d^2y}{dx^2}\) + 4y = 0
Substitute the equation (i) and equation (iii) in given differential equation,
\( \dfrac{ d^2y}{dx^2}\) + 4y = 0
-4 [A cos 2x – B sin 2x] + 4 (A cos 2x – B sin 2x) = 0
-4A cos 2x + 4B sin 2x + 4A cos 2x – 4B sin 2x = 0
0 = 0
Hence it is verified that, y = A cos 2x – B sin 2x is a solution of the differential equation is \( \dfrac{ d^2y}{dx^2}\)+ 4y = 0.
Answered by Aaryan | 1 year agoFind the particular solution satisfying the given condition \( (\dfrac{dy}{dx}) = (x – 1)\dfrac{dy}{dx} = 2x^3y\)
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