To find the general solution of the differential equation y⁴ + 2y = 0, we can start by factoring the equation.
First, we can factor out y from the equation:
y(y³ + 2) = 0
This gives us two cases to analyze:
Case 1: y = 0
This is one solution to the differential equation.
Case 2: y³ + 2 = 0
To solve y³ + 2 = 0, we can rearrange the equation:
y³ = -2
Taking the cube root of both sides gives:
y = -2^{1/3}
So, we have the other solution as well, which can be expressed in terms of imaginary numbers since the cube root of a negative number introduces complex solutions. The complex solutions can be found using Euler’s formula or directly as:
y = ext{cis}( rac{ heta + 2k heta}{3}) ext{ where } heta = an^{-1}(- rac{2}{ ext{current root}})
Overall, the general solution to the given differential equation thus combines:
- The constant solution y = 0
- The complex roots derived from the cubic equation y³ + 2 = 0
In summary, the general solution includes the trivial solution and any cubic roots derived from the non-real aspects of y³ + 2 = 0.