What is the general solution of the differential equation \[y’’ + 4y = 0\] ?

Hint: In this question, we need to find the general solution of the given differential equation \[y’’ + 4y = 0\] . The term differentiation is nothing but it is a process of determining the derivative of a function at any point . Mathematically, a derivative is defined as a rate of change of function with respect to an independent variable given in the function. The general solution of a linear equation if second order with constant coefficient consists of two parts namely the complementary function and the particular integral. The general solution of the differential equation is the sum of the particular integral and the complementary function. First, we need to find the characteristics equation . Then with roots of the characteristics equation, we can find the complementary function. Then finally we can find the general solution of the given differential equation.
Differential operators :
1.\[y = D = \dfrac{d}{{dx}}\]
2.\[y^{‘’} = D^{2} = \dfrac{d^{2}}{dx^{2}}\]
3.\[Dy = \dfrac{{dy}}{{dx}}\]

Complete step-by-step solution:
Given, \[y’’ + 4y = 0\]
We can rewrite the given equation as \[D^{2} + 4 = 0\]
General solution :
\[y = C.F + P.I\]
In order to find the complementary function, we need to solve the characteristics equation.
The characteristics equation is
\[p^{2} + 4 = 0\]
By taking \[p\] common,
We get,
\[p^{2} = - 4\]
Thus, \[p = \pm \sqrt{- 4}\]
On simplifying,
We get,
\[\Rightarrow \ p = \pm \sqrt{\left( - 1 \right)\left( 4 \right)}\]
We know that \[i^{2} = - 1\]
Thus \[p = \pm 2i\]
Thus we get \[p = 0 + 2i\] and \[p = 0 – 2i\]
The roots of the equation are complex roots \[m = p \pm iq\] . Therefore the complementary function is \[e^{{px}}\left({Acos}\left( {qx} \right) + Bcos\left( {qx} \right) \right)\] Where \[A\] and \[B\] are arbitrary constants.
Here \[p = 0\] and \[q = 2\]
Thus the complementary function is \[e^{0x}(Acos(2x) + B\cos\left( 2x \right))\].
Where \[e^{0}\] is \[1\] thus we get the complementary function is \[Acos(2x) + Bcos(2x)\] where \[A\] and \[B\] are arbitrary constants.
Since the given equation is in the form of \[a{_0}p^{2}\ + a{_1} p = 0\] , the particular integral is \[0\] .
Thus the general solution is \[y = 0 + Acos(2x) + Bcos(2x)\]
Therefore the general solution is \[y = Acos(2x) + Bcos(2x)\]
Final answer :
The general solution is \[y = Acos(2x) + Bcos(2x)\]

Note: Mathematically , There are two types of derivative namely first order derivative and second order derivative. The second order differential equation is nothing but the equation includes a second order derivative. When the roots are real and distinct with roots \[m = \alpha,\ \beta,\ldots\] which leads linearly independent solutions of the form \[y{_1} = e^{{\alpha x}}\] and \[y{_2}= e^{{\beta x}}\] . When the roots are real and repeated with the roots \[m = \alpha\] which leads a solution of the form \[y = \left( Ax + B \right)e^{{\alpha x}}\] in which the polynomial has the same degree . When the roots are complex roots in the form \[m = p \pm iq\] which leads a pairs linearly independent solutions of the form \[y = e^{{px}}\left( {Acos}\left( {qx} \right) + Bcos\left( {qx} \right) \right)\] .