We consider y 1 = sin 3 x y_1=\sin 3x y 1 = sin 3 x and y 2 = cos 3 x y_2=\cos 3x y 2 = cos 3 x . Then the requested wronskian is:
W ( y 1 , y 2 ) = ∣ y 1 y 2 y 1 ′ y 2 ′ ∣ = y 1 y 2 ′ − y 1 ′ y 2 W(y_1,y_2)=\begin{vmatrix}
y_1 & y_2\\
y_1' & y_2'
\end{vmatrix}=y_1y_2'-y_1'y_2 W ( y 1 , y 2 ) = ∣ ∣ y 1 y 1 ′ y 2 y 2 ′ ∣ ∣ = y 1 y 2 ′ − y 1 ′ y 2
y 1 = sin 3 x ⇒ y 1 ′ = 3 cos 3 x y 2 = cos 3 x ⇒ y 2 ′ = − 3 sin 3 x } \left.
\begin{array}{l}
y_1=\sin 3x\Rightarrow y_1'=3\cos 3x\\
y_2=\cos 3x\Rightarrow y_2'=-3\sin 3x
\end{array}
\right\} y 1 = sin 3 x ⇒ y 1 ′ = 3 cos 3 x y 2 = cos 3 x ⇒ y 2 ′ = − 3 sin 3 x } Substituting the latter into the Wronskian, we obtain:
W ( sin 3 x , cos 3 x ) = ∣ sin 3 x cos 3 x 3 cos 3 x − 3 sin 3 x ∣ = − 3 sin 2 3 x + 3 cos 2 3 x = W(\sin 3x, \cos 3x)=\begin{vmatrix}
\sin 3x & \cos 3x\\
3\cos 3x & -3\sin 3x
\end{vmatrix}=-3\sin^23x+3\cos^2 3x= W ( sin 3 x , cos 3 x ) = ∣ ∣ sin 3 x 3 cos 3 x cos 3 x − 3 sin 3 x ∣ ∣ = − 3 sin 2 3 x + 3 cos 2 3 x =
= 3 ( cos 2 3 x − sin 2 3 x ) = 3 cos ( 2 ⋅ 3 x ) = 3 cos 6 x =3(\cos^2 3x-\sin^2 3x)=3\cos (2\cdot 3x)=\boxed{3\cos 6x} = 3 ( cos 2 3 x − sin 2 3 x ) = 3 cos ( 2 ⋅ 3 x ) = 3 cos 6 x
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