Question

Find the general solution of the differential equation y''+y=0​

Answer 1

я написал решение на листочке ===>>

Answer 2

To find the general solution of the differential equation y'' + y = 0, we can assume a solution of the form y = e^(rt), where r is a constant.

Substituting this assumed solution into the differential equation, we get:

y'' + y = 0

(e^(rt))'' + e^(rt) = 0

r^2e^(rt) + e^(rt) = 0

Now, we can factor out e^(rt):

e^(rt)(r^2 + 1) = 0

Since e^(rt) is never zero, we must have (r^2 + 1) = 0. Solving this quadratic equation, we find two possible values for r:

r^2 + 1 = 0 r^2 = -1 r = ± i

Therefore, the general solution of the differential equation y'' + y = 0 is given by:

y = c1e^(it) + c2e^(-it)

Using Euler's formula, e^(ix) = cos(x) + i*sin(x), we can rewrite the general solution as:

y = c1(cos(t) + isin(t)) + c2(cos(t) - isin(t))

Simplifying further, we obtain:

y = (c1 + c2)cos(t) + (c1 - c2)i*sin(t)

Finally, we can rewrite the solution in terms of real functions:

y = Acos(t) + Bsin(t)

where A = c1 + c2 and B = (c1 - c2)i. A and B are arbitrary constants that depend on the initial conditions of the differential equation.