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Question: If \({\text{ }}y = 5\cos x - 3\sin x\) then, prove that \(\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0\)....

If  y=5cos⁡x−3sin⁡x{\text{ }}y = 5\cos x - 3\sin x then, prove that d2ydx2+y=0\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0.

Explanation

Solution

Hint : Differentiate the given equation twice then transpose to get the answer.

The given equation is ,
y=5cos⁡x−3sin⁡x ..............(i)   y = 5\cos x - 3\sin x{\text{ }}..............({\text{i}}) \\\ \\\
On differentiating the above equation we get,
dydx = −5sin⁡x−3cos⁡x ..............(ii)\dfrac{{dy}}{{dx}}{\text{ = }} - 5\sin x - 3\cos x{\text{ }}..............{\text{(ii)}}
Differentiating equation (ii) with respect to xx we get,
d2ydx2=−5cos⁡x+3sin⁡x\dfrac{{{d^2}y}}{{d{x^2}}} = - 5\cos x + 3\sin x
We can also write above equation as,
d2ydx2+5cos⁡x−sin⁡x=0\dfrac{{{d^2}y}}{{d{x^2}}} + 5\cos x - \sin x = 0
5cos⁡x−sin⁡x=y5\cos x - \sin x = y (From (i))
So,
d2ydx2+y=0\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0
Hence proved.

Note :- Whenever we are struck with these types of problems of proof that with the help of differentiation then we just have to differentiate the number of times the highest degree is present in the equation and then try to transpose something to get the required equation. Alternatively we can also differentiate and get the value separately and then put the values of the given equation .