Question

A third-order homogeneous linear equation and three linearly independent solutions are given. Find a particular solution satisfying the given initial conditions.y^(3) - 3y'' + 3y' - y = 0; y(0) = 2, y'(0) = 0, y'''(0) = 0; y1 = e^x, y2 = x e^x, y3 = x^2 e^x

Answer 1

You're given the general solution,

$y=C_1e^x+C_2xe^x+C_3x^2e^x$

Use the given initial values to find $C_1,C_2,C_3$:

$y(0)=2\implies2=C_1$

$y'=C_1e^x+C_2(1+x)e^x+C_3(2x+x^2)e^x$

$y'(0)=0\implies0=C_1+C_2$

$\implies C_2=-2$

$y''=C_1e^x+C_2(2+x)e^x+C_3(2+4x+x^2)e^x$

$y''(0)=0\implies0=C_1+2C_2+2C_3$

$\implies C_3=1$

Then the particular solution is

$y=2e^x-2xe^x+x^2e^x$