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Asked by Nasreen Last Modified
Answer To verify that the function y=acos(x)+bsin(x)y=acos(x)+bsin(x), where a,b∈Ra,b∈R, is a solution of the differential equation d2ydx2+y=0dx2d2y+y=0, we need to substitute this function into the given differential equation and show that it satisfies the equation.
Given function: y=acos(x)+bsin(x)y=acos(x)+bsin(x)
First, let's find the first and second derivatives of yy with respect to xx:
dydx=−asin(x)+bcos(x)dxdy=−asin(x)+bcos(x) d2ydx2=−acos(x)−bsin(x)dx2d2y=−acos(x)−bsin(x)
Now, let's substitute these derivatives into the given differential equation:
d2ydx2+y=(−acos(x)−bsin(x))+(acos(x)+bsin(x))dx2d2y+y=(−acos(x)−bsin(x))+(acos(x)+bsin(x))
=(−acos(x)+acos(x))+(−bsin(x)+bsin(x))=(−acos(x)+acos(x))+(−bsin(x)+bsin(x))
=0=0
Since the expression simplifies to 0, it verifies that the function y=acos(x)+bsin(x)y=acos(x)+bsin(x) is a solution of the differential equation d2ydx2+y=0dx2d2y+y=0.
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