This question was also an issue for me Any help would be muc

This question was also an issue for me. Any help would be much appreciated.

Solve the given differential equation by undetermined coefficients and the annihilator method.

y\'\'+4y\'+4y=5x+7

Thank you for your time.

Solution

y\'\' + 4y\' + 4y = 5x + 7

First we need to write the characteristic equation, for characteristic equation y\'\' coefficient will come with p^2, y\' with p and y term will come as it is

characteristics equation of above polynomial

p^2 + 4p + 4 = 0

(p+2)^2 = 0

since in this case both the roots are -2 and -2

Case : If both the roots are equal in order to be linearly independent, we need to multiply one root by x

Hence yh(x) = c1*e^(-2x) + c2*x*e^(-2x)

Now we need to solve for the yp (particular solution), let us assume the particular solution is of the form of

yp = Ax^2 + Bx + C

yp\' = 2Ax + B

yp\'\' = 2A

2A + 4(2Ax + B) + 4(Ax^2 + Bx + C) = 5x + 7

4Ax^2 + (8A + 4B)x + (2A+4B+4C) = 5x + 7

Hence A = 0 since there is no coefficient of x^2

4B = 5 or B = 5/4

and 4B +4C = 7

C = 1/2

Hence yp(x) = 5/4x + 1/2

Final solution is y = yh + yp = c1*e^(-2x) + c2*x*e^(-2x) + 5/4x + 1/2