Hard problem on differential equation problem?

1. y'''' + 10y'' + 12y = 8 sin 2x + 32 cos 4x

(1) Homogeneous Solution

Operator function: L(D) = D^4 + 10D^2 + 12 = 0

D^2 = -5 + - √13

D = +- √(-5 + - √13) = +-a*i; +-b*i

where i^2 = -1

a = +- √|-5 + √13|

b = +- √|-5 - √13|


yh(x) = c1*cos(ax) + c2*sin(ax) + c3*cos(bx) + c4*sin(bx)



(2) Particular Solution

L(-i2^2) = L(-4i) = 64 - 10 * 16 + 12 = -84

L(-i4^2) = L(-16i) = 65536 - 10*256 + 12 = 62988

sin(2x)/L(-4i) = sin(2x)/84

cos(4x)/L(-16i) = cos(4x)/62988


Using inverse operators, it results in

yp(x) = -sin(2x)/84 + cos(4x)/62988



(3) General Solution

y(x) = c1*cos(ax) + c2*sin(ax) + c3*cos(bx) + c4*sin(bx)

- sin(2x)/84 + cos(4x)/62988






2. 0 = y'' + e ^4y (y')^6

= (y')' + 0.25*[exp(4y)]'*(y')^5 ... Note1

= (y')'/(y')^5 + 0.25*[e^(4y)]'

= -0.25*[1/(y')^4]' + 0.25*[e^(4y)]' ... Note2

= -d[1/(y')^4]/dx + d[e^(4y)]dx

c = -∫d[1/(y')^4] + ∫d[e^(4y)]



It can be integrated by setting c = 0

(y')^4 = 1/(e^y)^4

y' = 1/e^y

∫dx = ∫e^y*dy


(x + c1) = e^y

y(x) = ln|x + c1| = Answer


Note1. [e^(4y)]' = 4y'*e^(4y)

Note2.(1/y'^4)' = -4y"/y'^5