The Differential Equation formed by y = across + bsinx + 4, where a and b are arbitrary constant is?
The Differential Equation formed by y = across + bsinx + 4, where a and b are arbitrary constant is?
Explanation
Given y = acos(x) + bsin(x) + 4
First derivative: dy/dx = -asin(x) + bcos(x)
Second derivative: d²y/dx² = -acos(x) - bsin(x)
Notice that d²y/dx² = -(acos(x) + bsin(x))
Given y = acos(x) + bsin(x) + 4, we rearrange to find acos(x) + bsin(x) = y - 4
Substituting into the second derivative equation gives:
d²y/dx² = -(y - 4)
d²y/dx² = -y + 4
d²y/dx² + y = 4
The answer is d²y/dx² + y = 4.