# What is the general solution of the differential equation y'' + 4y =0?

**Solution:**

y'' + 4y = 0 [Given]

The characteristic equation can be written as

r^{2} + 4 = 0

The formula for the standard form of quadratic equation ax^{2} + bx + c = 0 is written as

x = [−b ± √(b^{2} − 4ac)] / 2a

From the given equation we know that

a = 1

b = 0

c = 4

Substituting it in the formula

x = [−0 ± √{(0)^{2} − 4 × 1 × 4}] / 2(1)

x = [-0 ± √0 − 16] / 2

By further calculation

x = ±2i …. (1)

The general solution of a second order with complex roots v ± wi is written as

y = e^{v}x [C cos (wx) + iD sin (wx)] …. (2)

From the equation (1),

substitute v = 0 and w = 2 in (2)

y = e^{0}x [ C cos (2x) + iD sin (2x) ]

Therefore, the general solution is y = [ C cos (2x) + iD sin (2x) ].

## What is the general solution of the differential equation y'' + 4y =0?

**Summary: **

The general solution of the differential equation y'' + 4y = 0 is y = [C cos (2x) + I D sin (2x) ].

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