Question

Pls solve (D^3-3D+2)y=e^5x

Answer 1

Ptolemy is wrong because the homogeneous equation has the general solution y = Ae^(-x) + Be^(-2x). Because the nonhomogeneous function is e^(e^x), we cannot use the method of undetermined coefficients; we must use variation of parameters. 

We seek functions E(x) and F(x) such that 
E(x)e^(-x) + F(x)e^(-2x) is a particular solution of the nonhomogeneous equation. We find that the derivatives of E and F must satisfy the conditions 

E'(x)e^(-x) + F'(x)e^(-2x) = 0 
-E'(x)e^(-x) - 2F'(x)e^(-2x) = e^(e^x). 

We simplify these conditions by multiplying through by e^x: 

E' + F'e^(-x) = 0 
-E' - 2F'e^(-x) = e^(e^x) 

We find that E'(x) = e^(e^x + x) and F'(x) = - e^(e^x + 2x). Taking the antiderivatives, we find E(x) = e^(e^x) 
and F(x) = e^(e^x) - (e^x)*e^(e^x). 

Now E*e^(-x) + F*e^(-2x) = e^(e^x - 2x); you can verify that this is a particular solution. The general solution is 
y = Ae^(-x) + Be^(-2x) + e^(e^x - 2x). 

Answer 2

Answer:

Simplifying

(D2 + -3D + 2) * y = e5x

Reorder the terms:

(2 + -3D + D2) * y = e5x

Reorder the terms for easier multiplication:

y(2 + -3D + D2) = e5x

(2 * y + -3D * y + D2 * y) = e5x

(2y + -3yD + yD2) = e5x

Solving

2y + -3yD + yD2 = e5x

Solving for variable 'y'.

Move all terms containing y to the left, all other terms to the right.

Reorder the terms:

-1e5x + 2y + -3yD + yD2 = e5x + -1e5x

Combine like terms: e5x + -1e5x = 0

-1e5x + 2y + -3yD + yD2 = 0

The solution to this equation could not be determined.

Step-by-step explanation: