Question

Solve y''-4y'+4y=xe^(2x)

complete this question using the table find y' and y'' to sub in original to get yp and yn=yh+yp
y''-4y+4y=xe^(3)
yh=C1*e^(2x)+C2*xe^(2x)
yp=(e^(2x))(Ax+B)x^2
yp=e^(2x)(Ax^3+Bx^2)

Answer 1

To solve the differential equation y'' - 4y' + 4y = xe^(2x), we can find the particular solution and the complementary solution separately. The complementary solution (yh) is given by yh = C1e^(2x) + C2xe^(2x), where C1 and C2 are arbitrary constants. To find the particular solution (yp), we can use the method of undetermined coefficients by guessing that yp = (e^(2x))(Ax + B)x^2 and solving for A and B.

Step-by-step explanation:

To solve the differential equation y'' - 4y' + 4y = xe^(2x), we can find the particular solution and the complementary solution separately.

The complementary solution (yh) is given by yh = C1e^(2x) + C2xe^(2x), where C1 and C2 are arbitrary constants.

To find the particular solution (yp), we can use the method of undetermined coefficients. By guessing that yp = (e^(2x))(Ax + B)x^2, we can substitute it into the original differential equation and solve for A and B.