Differential Equation

Find the general solution of the following differential equation:
y’”(t) = 4cos t 3 sin 2t
Solution
To find the general solution of the given third-order linear differential equation:
y”'(t) = 4cos(t) 3sin(2t)
We will need to solve the differential equation by integrating it multiple times. Let’s proceed step-by-step:
Step 1: First Integration To find the solution for y”(t), we integrate both sides of the equation with respect to t:
∫ y”'(t) dt = ∫ (4cos(t) 3sin(2t)) dt
y”(t) = 4∫ cos(t) dt 3∫ sin(2t) dt
y”(t) = 4sin(t) – (3/2)cos(2t) C1 (where C1 is the constant of integration)
Step 2: Second Integration Next, we integrate y”(t) to find the solution for y'(t):
∫ y”(t) dt = ∫ (4sin(t) – (3/2)cos(2t) C1) dt
y'(t) = -4∫ sin(t) dt – (3/2)∫ cos(2t) dt C1t C2 (where C2 is the constant of integration)
y'(t) = 4cos(t) – (3/4)sin(2t) C1t C2
Step 3: Third Integration Finally, we integrate y'(t) to find the general solution for y(t):
∫ y'(t) dt = ∫ (4cos(t) – (3/4)sin(2t) C1t C2) dt
y(t) = 4∫ cos(t) dt – (3/4)∫ sin(2t) dt ∫ C1t dt ∫ C2 dt
y(t) = 4sin(t) (3/8)cos(2t) (C1/2)t^2 C2t C3 (where C3 is the constant of integration)
So, the general solution to the given third-order linear differential equation is:
y(t) = 4sin(t) (3/8)cos(2t) (C1/2)t^2 C2t C3
where C1, C2, and C3 are arbitrary constants that can be determined based on initial conditions if they are given.