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Question: The differential equation whose solution is \(y = c _ { 1 } \cos a x + c _ { 2 } \sin a x\) is (Wh...

The differential equation whose solution is

y=c1cosax+c2sinaxy = c _ { 1 } \cos a x + c _ { 2 } \sin a x is

(Where c1,c2c _ { 1 } , c _ { 2 } are arbitrary constants)

A

d2ydx2+y2=0\frac { d ^ { 2 } y } { d x ^ { 2 } } + y ^ { 2 } = 0

B

d2ydx2+a2y=0\frac { d ^ { 2 } y } { d x ^ { 2 } } + a ^ { 2 } y = 0

C

d2ydx2+ay2=0\frac { d ^ { 2 } y } { d x ^ { 2 } } + a y ^ { 2 } = 0

D

d2ydx2a2y=0\frac { d ^ { 2 } y } { d x ^ { 2 } } - a ^ { 2 } y = 0

Answer

d2ydx2+a2y=0\frac { d ^ { 2 } y } { d x ^ { 2 } } + a ^ { 2 } y = 0

Explanation

Solution

Solution is y=c1cosax+c2sinaxy = c _ { 1 } \cos a x + c _ { 2 } \sin a x

Differentiate it w.r.t. x, we get

dydx=c1asinax+c2acosax\frac { d y } { d x } = - c _ { 1 } a \sin a x + c _ { 2 } a \cos a x

Again d2ydx2=c1a2cosaxc2a2sinax\frac { d ^ { 2 } y } { d x ^ { 2 } } = - c _ { 1 } a ^ { 2 } \cos a x - c _ { 2 } a ^ { 2 } \sin a x

d2ydx2=a2(c1cosax+c2sinax)d2ydx2=a2y\frac { d ^ { 2 } y } { d x ^ { 2 } } = - a ^ { 2 } \left( c _ { 1 } \cos a x + c _ { 2 } \sin a x \right) \Rightarrow \frac { d ^ { 2 } y } { d x ^ { 2 } } = - a ^ { 2 } y

or d2ydx2+a2y=0\frac { d ^ { 2 } y } { d x ^ { 2 } } + a ^ { 2 } y = 0.