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.