can someone help me find the general solution to
d^2*y/dx^2 - 2dy/dx + y = x^2*e^x
if you can also go through in detail, it would be much appreciated =D
d^2*y/dx^2 - 2dy/dx + y = x^2*e^x
if you can also go through in detail, it would be much appreciated =D
-
m² - 2m + 1 = 0
(m - 1)² = 0
m = 1, 1.
yc = c1e^x + c2xe^x.
_______________________
W(x) = | e^x xe^x|
|e^x (x+1)e^x|
= (x+1)e^(2x) - xe^(2x)
= e^2x.
_______________________
yp = -y1∫y2(f(x))/[W(x)a(x)]dx + y2∫y1(f(x))/[W(x)a(x)]dx
Where a(x) is the coefficient of the y'' term (in this case a(x) = 1).
W(x) is the wronskian of y1, y2. And f(x) is the right hand side of the equation (x²e^x).
_______________________
yp = -e^x∫(xe^x)(x²e^x)/[e^(2x)]dx + xe^x∫e^x(x²e^x)/[e^(2x)]dx
yp = -e^x∫(x)(x²)dx + xe^x∫(x²)dx
yp = -e^x∫x³dx + xe^x∫(x²)dx
yp = -e^x[1/4 x⁴] + xe^x[⅓x³]
yp = -1/4 e^x[x⁴] + ⅓ x⁴ e^x
yp = (⅓ -1/4)x⁴ e^x
yp = (4/12 -3/12)x⁴ e^x
yp = (1/12)x⁴ e^x
___________________________________
y = yc + yp
y = c1e^x + c2xe^x + (1/12)x⁴ e^x
(m - 1)² = 0
m = 1, 1.
yc = c1e^x + c2xe^x.
_______________________
W(x) = | e^x xe^x|
|e^x (x+1)e^x|
= (x+1)e^(2x) - xe^(2x)
= e^2x.
_______________________
yp = -y1∫y2(f(x))/[W(x)a(x)]dx + y2∫y1(f(x))/[W(x)a(x)]dx
Where a(x) is the coefficient of the y'' term (in this case a(x) = 1).
W(x) is the wronskian of y1, y2. And f(x) is the right hand side of the equation (x²e^x).
_______________________
yp = -e^x∫(xe^x)(x²e^x)/[e^(2x)]dx + xe^x∫e^x(x²e^x)/[e^(2x)]dx
yp = -e^x∫(x)(x²)dx + xe^x∫(x²)dx
yp = -e^x∫x³dx + xe^x∫(x²)dx
yp = -e^x[1/4 x⁴] + xe^x[⅓x³]
yp = -1/4 e^x[x⁴] + ⅓ x⁴ e^x
yp = (⅓ -1/4)x⁴ e^x
yp = (4/12 -3/12)x⁴ e^x
yp = (1/12)x⁴ e^x
___________________________________
y = yc + yp
y = c1e^x + c2xe^x + (1/12)x⁴ e^x