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Asked by Neetu Last Modified
Answer To form the differential equation representing the family of curves y=asin(x+b)y=asin(x+b), where aa and bb are arbitrary constants, we need to find the first and second derivatives of yy with respect to xx.
Given function: y=asin(x+b)y=asin(x+b)
First, let's find the first derivative of yy with respect to xx:
dydx=acos(x+b)dxdy=acos(x+b)
Now, let's find the second derivative of yy with respect to xx:
d2ydx2=−asin(x+b)dx2d2y=−asin(x+b)
Now, we need to express yy, dydxdxdy, and d2ydx2dx2d2y in terms of yy and xx to form the differential equation.
We have:
y=asin(x+b)y=asin(x+b)
dydx=acos(x+b)dxdy=acos(x+b)
d2ydx2=−asin(x+b)dx2d2y=−asin(x+b)
Substituting these into the differential equation d2ydx2+y=0dx2d2y+y=0, we get:
−asin(x+b)+asin(x+b)=0−asin(x+b)+asin(x+b)=0
0=00=0
This equation is trivially satisfied.
However, let's represent the differential equation without cancelling out terms:
d2ydx2+y=0dx2d2y+y=0
−asin(x+b)+asin(x+b)+asin(x+b)=0−asin(x+b)+asin(x+b)+asin(x+b)=0
asin(x+b)−asin(x+b)=0asin(x+b)−asin(x+b)=0
0=00=0
This differential equation represents the family of curves y=asin(x+b)y=asin(x+b), where aa and bb are arbitrary constants.
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