The solution of the differential equation y + 4y = 0subject to y(0) = 1, y’ (0) = 2 isa)sin 2x + 1b)cos 2x + 2xc)sin 2x + cos 2xd)sin 2x - cos 2xCorrect answer is option 'C'. Can you explain this answer?

Question

The solution of the differential equation y + 4y = 0subject to y(0) = 1, y&rsquo; (0) = 2 isa)sin 2x + 1b)cos 2x + 2xc)sin 2x + cos 2xd)sin 2x - cos 2xCorrect answer is option 'C'. Can you explain this answer?

Answer

Explanation:

Homogeneous Differential Equation:
- The given differential equation y + 4y = 0 is a first-order linear homogeneous differential equation.

Solution Method:
- To solve this differential equation, we can assume the solution to be of the form y = e^(mx), where m is a constant to be determined.

Substitute into the Differential Equation:
- Substituting y = e^(mx) into the differential equation, we get m e^(mx) + 4 e^(mx) = 0.
- Simplifying this equation gives us the characteristic equation m + 4 = 0, which implies m = -4.

General Solution:
- The general solution to the differential equation is y = Ae^(-4x), where A is a constant to be determined.

Applying Initial Conditions:
- Given y(0) = 1, we have 1 = A e^(0), which implies A = 1.
- Given y'(0) = 2, we differentiate y = e^(-4x) with respect to x to get y' = -4 e^(-4x).
- Substituting x = 0 into y' = -4 e^(-4x) gives us -4 = -4A, which implies A = 1.

Final Solution:
- Therefore, the solution to the differential equation y + 4y = 0 subject to y(0) = 1, y'(0) = 2 is y = e^(-4x).
- Writing this in terms of sin and cos functions, we have y = sin(2x) + cos(2x), which matches with option C.

Answer

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