Solve : ( D² + 3D + 2) y = e^(e^x)?

Question

Answer : y = ae^(-x) + be^(-2x) + ce^(-3x) + e^( e^x - 2x )
its urgent!
please show the full procedure.

I have provided the answer.

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).

I don't know where you get the term ce^(-3x). A second order equation only has two arbitrary constants. Moreover, e^(-3x) does not satisfy the homogeneous equation.

Answer 2

Rewrite this differential equation in well-known kind: (D² - 3-D² + 4D - 2)y = ?? (4D - 2nd² - 2)y = ?? 4y' - 2y'' - 2y = ?? discover the complementary function by skill of fixing the auxiliary equation: 4m - 2m² - 2 = 0 2m² - 4m + 2 = 0 m² - 2m + a million = 0 (m - a million)² = 0 m - a million = 0 m = a million y? = (C?x + C?)?? discover the specific indispensable by skill of evaluating coefficients: y? = Ax²?? y?' = (Ax² + 2Ax)?? y?'' = (Ax² + 4Ax + 2A) 4y?' - 2y?'' - 2y? = ?? 4(Ax² + 2Ax)?? - 2(Ax² + 4Ax + 2A) - 2Ax²?? = ?? (4Ax² + 8Ax)?? - (2Ax² + 8Ax + 4A)?? - 2Ax²?? = ?? -4A?? = ?? 4A = -a million A = -¼ y? = -x²?? / 4 discover the final answer by skill of combining those 2 aspects: y = y? + y? y = (C?x + C?)?? - x²?? / 4 y = (C?x - x² + C?)?? / 4

Answer 3

( D² + 3D + 2) y = e^(e^x)

solve for complementary solution
( D² + 3D + 2) y =0
m^2+3m+2=0
(m+2)(m+1)=0
m=-2,m=1
yc=c1exp(-2t)+c2exp(t)

Solve for yp
D(yc)=-2c1exp(-2t)+c2exp(t)
D^2(yc)=4c1exp(-2t)+c2exp(t)

substitute the derivative of yc to the given equation

-2c1exp(-2t)+c2exp(t)+3[4c1exp(-2t)+c2exp(t)4c1exp(-2t)+c2exp(t)]+
2[c1exp(-2t)+c2exp(t)]=e^(e^t)