Answer:
a) To solve the differential equation y''-2y'-3y= e^4x, we first find the characteristic equation:
r^2 - 2r - 3 = 0
Factoring, we get:
(r - 3)(r + 1) = 0
So the roots are r = 3 and r = -1.
The general solution to the homogeneous equation y'' - 2y' - 3y = 0 is:
y_h = c1e^3x + c2e^(-x)
To find the particular solution, we use the method of undetermined coefficients. Since e^4x is a solution to the homogeneous equation, we try a particular solution of the form:
y_p = Ae^4x
Taking the first and second derivatives of y_p, we get:
y_p' = 4Ae^4x
y_p'' = 16Ae^4x
Substituting these into the original differential equation, we get:
16Ae^4x - 8Ae^4x - 3Ae^4x = e^4x
Simplifying, we get:
5Ae^4x = e^4x
So:
A = 1/5
Therefore, the particular solution is:
y_p = (1/5)*e^4x
The general solution to the non-homogeneous equation is:
y = y_h + y_p
y = c1e^3x + c2e^(-x) + (1/5)*e^4x
b) To solve the differential equation y'' + y' - 2y = 3xe^x, we first find the characteristic equation:
r^2 + r - 2 = 0
Factoring, we get:
(r + 2)(r - 1) = 0
So the roots are r = -2 and r = 1.
The general solution to the homogeneous equation y'' + y' - 2y = 0 is:
y_h = c1e^(-2x) + c2e^x
To find the particular solution, we use the method of undetermined coefficients. Since 3xe^x is a solution to the homogeneous equation, we try a particular solution of the form:
y_p = (Ax + B)e^x
Taking the first and second derivatives of y_p, we get:
y_p' = Ae^x + (Ax + B)e^x
y_p'' = 2Ae^x + (Ax + B)e^x
Substituting these into the original differential equation, we get:
2Ae^x + (Ax + B)e^x + Ae^x + (Ax + B)e^x - 2(Ax + B)e^x = 3xe^x
Simplifying, we get:
3Ae^x = 3xe^x
So:
A = 1
Therefore, the particular solution is:
y_p = (x + B)e^x
Taking the derivative of y_p, we get:
y_p' = (x + 2 + B)e^x
Substituting back into the original differential equation, we get:
(x + 2 + B)e^x + (x + B)e^x - 2(x + B)e^x = 3xe^x
Simplifying, we get:
-xe^x - Be^x = 0
So:
B = -x
Therefore, the particular solution is:
y_p = xe^x
The general solution to the non-homogeneous equation is:
y = y_h + y_p
y = c1e^(-2x) + c2e^x + xe^x
c) To solve the differential equation y" - 9y' + 20y = x^2*e^4x, we first find the characteristic equation:
r^2 - 9r + 20 = 0
Factoring, we get:
(r - 5)(r - 4) = 0
So the roots are r = 5 and r = 4.
The general solution to the homogeneous equation y" - 9y' + 20y = 0 is:
y_h = c1e^4x + c2e^5x
To find the particular solution, we use the method of undetermined coefficients. Since x^2*e^4x is a solution to the homogeneous equation, we try a particular solution of the form:
y_p = (Ax^2 + Bx + C)e^4x
Taking the first and second derivatives of y_p, we get:
y_p' = (2Ax + B)e^4x + 4Axe^4x
y_p'' = 2Ae^4x +