Calculus Question?

First, find the derivative of y.

y=a sin bx
y'=ab cos bx
y''= -ab^2sinbx

Note that y' is the same as dy/dx and y'' is d^2y/dx^2, it's just a different way of writing it.

Now substitute the equations into (d^2y/dx^2)+b^2y=0

-ab^2sinbx + b^2 * a sin bx

and that should add up to be 0

dy/dx=ab cos bx;
d^2 y/dx^2=-ab^2 sin bx;
b^2 y=ab^2 sin bx


d^2y/dx^2+b^2 y=
-ab^2 sin bx+ab^2 sin bx=0

y = a sin(bx)
dy/dx = a cos(bx) * b = ab cos(bx)
d²y/dx² = ab (-sin(bx)) * b = -ab² sin(bx)

d²y/dx² + b²y = -ab² sin(bx) + b² a sin(bx) = 0

Take the double derivative of y with respect to x and plug it in above to see if it gives you zero. That's all it is asking to do.

This is separable:

d²y/dx² + b²y = 0
d²y/dx² = −b²y
(1/y) d²y/dx² = −b²
(1/y) d²y = −b² dx²
∫ ∫ (1/y) d²y = ∫ ∫ −b² dx²
∫ ln(y) + C dy = ∫ −b²x + S dx
y ⋅ [ ln(y) − 1 ] + Cy + K = −b²x² + Sx + T

Simplify:
y ⋅ [ ln(y) − 1 ] + Cy = −b²x² + Sx − K
y⋅ln(y) − y + Cy = −b²x² + Sx − K
y ⋅ [ ln(y) + C ] = −b²x² + Sx − K
y ⋅ [ ln(y) + ln(C) ] = −b²x² + Sx − K
y⋅ln(Cy) = −b²x² + Sx − K

let y = e^z and thus ln(y) = z
e^z ⋅ ln(Ce^z) = −b²x² + Sx − K
e^z ⋅ ln(e^C⋅e^z) = −b²x² + Sx − K
e^z ⋅ ln(e^{C + z}) = −b²x² + Sx − K
e^z ⋅ (C + z) = −b²x² + Sx − K

e^z ⋅ e^C ⋅ (C + z) = ( −b²x² + Sx − K ) ⋅ e^C
e^(C + z) ⋅ (C + z) = ( −b²x² + Sx − K ) ⋅ e^C

Apply Lambert W-Function:
C + z = W[ ( −b²x² + Sx − K ) ⋅ e^C ]
z = W[ ( −b²x² + Sx − K ) ⋅ e^C ] − C

Since ln(y) = z
ln(y) = W[ ( −b²x² + Sx − K ) ⋅ e^C ] − C
y = e^[ W[ ( −b²x² + Sx − K ) ⋅ e^C ] − C ]

Simplify:
y = e^W[ ( −b²x² + Sx − K ) ⋅ e^C ] / e^C
y = e^W( −Cb²x² + CSx − CK ) / C
y = e^W( Cb²x² + Sx + K ) / C

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Alternative approach:

d²y/dx² + b²y = 0
y'' + b²y = 0

characteristic equation
r² + b² = 0
r² = −b²
r = ±bi

Therefore:
y₁ = C⋅ e^(bi)
y₂ = K⋅ e^(−bi)

y₁ = C⋅ [ cos(b) + i⋅sin(b) ]
y₂ = K⋅ [ cos(−b) + i⋅sin(−b) ]

y₁ = C⋅ [ cos(b) + i⋅sin(b) ]
y₂ = K⋅ [ cos(b) − i⋅sin(b) ]

y_h = y₁ + y₂
y_h = C⋅ [ cos(b) + i⋅sin(b) ] + K⋅ [ cos(b) − i⋅sin(b) ]
y_h = C⋅ cos(b) + K⋅sin(b)


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So some douche decided to down thumb me and hide behind the anonymity of the internet. Coward. Come on... whoever you are, if you have a complaint or a whine, the least you could do is say so.

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Jane... I am just showing alternative methods to arrive at different answers. Educating those who ask.