Can't solve diff eq homework problem: y" = -3y' - 2y -et

bem777

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Directly quoted:
Consider the differential equation:
y" = -3y' - 2y -et


i) What sort of differential equation is this?

ii) What is yh(t), the general homogeneous solution?

iii) What is yp(t), the particular inhomogeneous solution?

iv) If y(0) = 0 and y(ln(2)) = 1, what is the full specific solution?


So far I have gotten:

Second Order Linear inhomogeneous differential equation

y" + 3y' + 2y = 0
<L = Lambda>
L 2 + 3L + 2 = 0
(L + 2)(L + 1) L = -2, -1
Yn(t) = c1 e-2t + c2 e
-t
y" + 3y' +2y = -et
y(t) = aect
y'(t) = acect
y"(t) = ac2ect

C = 1

a(1)2et + 3a(1)et + 2aet = -et
aet + 3aet + 2aet = -et
aet + 5aet + 2aet = -et
6aet = -et 6a = -1 a = -1/6


...and because I could see that this didn't make sense I just stopped. Can someone help me understand where I got off track and how I should have proceeded?
 
Directly quoted:
Consider the differential equation:
y" = -3y' - 2y -et


i) What sort of differential equation is this?

ii) What is yh(t), the general homogeneous solution?

iii) What is yp(t), the particular inhomogeneous solution?

iv) If y(0) = 0 and y(ln(2)) = 1, what is the full specific solution?


So far I have gotten:

Second Order Linear inhomogeneous differential equation

y" + 3y' + 2y = 0
<L = Lambda>
L 2 + 3L + 2 = 0
(L + 2)(L + 1) L = -2, -1
Yn(t) = c1 e-2t + c2e
-t
y" + 3y' +2y = -et
y(t) = aect
y'(t) = acect
y"(t) = ac2ect

C = 1

a(1)2et + 3a(1)et + 2aet = -et
aet + 3aet + 2aet = -et
aet + 5aet + 2aet = -et
6aet = -et 6a = -1 a = -1/6


...and because I could see that this didn't make sense I just stopped. Can someone help me understand where I got off track and how I should have proceeded?

Your particular solution should be:

yp = -1/6 * et

Then your complete solution is:

Y(t) = c1 e-2t + c2e-t - 1/6 * et

Now evaluate c1 & c2 using given bdy. conditions.
 
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