Solve this second order linear ordinary differential equation?

using homogeneous:

Solve ( d^2 y(x))/( dx^2)+y(x) = 0:

Assume a solution will be proportional to e^(lambda x) for some constant lambda.

Substitute y(x) = e^(lambda x) into the differential equation:

( d^2 )/( dx^2)(e^(lambda x))+e^(lambda x) = 0

Substitute ( d^2 )/( dx^2)(e^(lambda x)) = lambda^2 e^(lambda x):

lambda^2 e^(lambda x)+e^(lambda x) = 0

Factor out e^(lambda x):

(lambda^2+1) e^(lambda x) = 0

Since e^(lambda x) !=0 for any finite lambda, the zeros must come from the polynomial:

lambda^2+1 = 0

Solve for lambda:

lambda = i or lambda = -i

The roots lambda = ± i give y_1(x) = c_1 e^(i x), y_2(x) = c_2 e^(-i x) as solutions, where c_1 and c_2 are arbitrary constants.

The general solution is the sum of the above solutions:

y(x) = y_1(x)+y_2(x) = c_1 e^(i x)+c_2/e^(i x)

Apply Euler's identity e^(alpha+i beta) = e^alpha cos(beta)+i e^alpha sin(beta):

y(x) = c_1 (cos(x)+i sin(x))+c_2 (cos(x)-i sin(x))

Regroup terms:

y(x) = (c_1+c_2) cos(x)+i (c_1-c_2) sin(x)

Redefine c_1+c_2 as c_1 and i (c_1-c_2) as c_2, since these are arbitrary constants:

Answer: y(x) = c_1 cos(x)+c_2 sin(x)

Here's last few steps in image: