Answer :
The exercise above is one of the Discontinuous Initial Value Problems (DIVP) with a Discontinuous Function.
An Initial value problem
This topic is related to multivariable calculus. It is a standard differential equation with an initial condition that describes the unknown function's value at a selected point in the domain. IVP is often used in physics for modeling.
The solution for the IVP (Initial Value Problem) whose input function g(x) is discontinuous
y" + 4y = g(x),
y(0) =1, and
y' (0) = 2
Where
g(x) = [ sin x , 0 ≤ x ≤ π /2 ]
[ , x > / 2 ]
For the lower interval and use the result to solve the upper interval.
The homogeneous solution to (1) where (1) is
y′′ + 4y = g(x), y(0) = 1, y′(0) = 2
m² + 4 = 0 → m₁,₂ = ± 2i
From this derive:
y h = c₁ Cos 2x + c₂ sin 2x
To resolve the particular solution,
y p = a cos x + b sin x
Resolving the constants a and b, let's substitute back into equation 1 above to arrive at:
-a cos x - b sin x + 4a cos x + 4b sin x = sin x
From the above,
a = 0 and b= 1/3
Rewriting the equation,
y = y h + y p = c₁ cos 2x + c₂ sin 2x + 1/3sinx
Initiation constraints
In mathematics, a constraint is a condition of an optimization problem that the solution must satisfy. There are several types of constraints—primarily equality constraints, inequality constraints, and integer constraints. The set of candidate solutions that satisfy all constraints is called the feasible set.
Using the Initiation constraints, we solve for c₁ and c₂:
c₁ = 1, c₂ = 5/6, this leaves us with the following:
y(x) =cos2x + 5/6sin 2x + 1/3sinx,
0 ≤ x ≤ π / 2 (2)
From the above, we must proceed to calculate the upper half of the range so that we have, y" + 4y = 0.
Using the initial constraints in combination with (2), we get y = -(2/3) and
y' (/2) = -(5/3)
Recall that solution 2 was:
y = c₁ cos 2x + c₂ sin 2x
With the new constraints, we can resolve this get:
c₁ = 2/3, c₂ = 5/6, so
y(x) = 2/3cos 2x + 5/6sin 2x, x > /3 (3)
Combining equations 2 and 3, we have:
y(x) = [ cos 2x + 5/6 sin 2x + 1/3 sin x , 0 ≤ x ≤ π / 2 ]
= [ 2/3 cos 2x + 5/6 sin 2x, x > π /2 ]
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