Question

Find y as a function of x if
y^(4)−4y'''+4y''=128e^(−2x)
y(0)=9 y'(0)=6 y''(0)=12 y'''(0)=−16

Answer 1

This is a nonhomogeneous differential equation with constant equation. It has a solution of the form:
\[y(x)=y_c+y_p\]
where y_c is the complementary solution associated with its homogeneous equation, and y_p is the particular solution.

Answer 2

To find y_c, we just have to solve the auxiliary equation associated with the homogeneous equation, that's:
\[m^4-4m^3+4m^2=0 \implies m^2(m-2)^2=0\]
Which has double roots of 0 and 2. Then:
\[y_c=c_1+c_2x+c_3e^{2x}+c_4xe^{2x}\]

Answer 3

Now, let's assume a particular solution Ae^(-2x), substitute in the original differential equation:
\[16Ae^{-2x}-4(-8Ae^{-2x})+4(4Ae^{-2x})=128e^{-2x} \implies 16A+36A+16A=128\]
and hence A=2, therefore the particular solution is:
\[y_p=2e^{-2x}\]

Answer 4

Therefore the general solution is:
\[y=c_1+c_2x+c_3e^{2x}+c_4xe^{2x}+2e^{-2x}\]
Now, use the given initial conditions to find the constants c_1, c_2,c_3 and c_4.