The solution (up to three decimal places) at x = 1 of the differentia...
Given D.E. is
∴ The auxiliary equation is:
(m + 1)2 = 0
⇒ m = -1, -1 [real and repeated]
The complementary function (C.F.):
yC = e-x[c1 + c2x]
And
∴ yp = 0 [particular integral]
∴ The complete solution of (1) is:
y = e-x[c1 + c2x]→(2)
Given that y(0) = 1
From (1) 1 = c1 + 0 ⇒ c1 = 1
dy/dx = -e-x[c1 + c2x] + e-x[c2]
Given: dy/dx = -1 at x = 0
-1 = -[1 + 0] + 1[c2] ⇒ c2 = 0
∴ y = e-x [1 + 0] = e-x ⇒ y(1) = e-1 = 0.368
II method:
Given D.E. is:
⇒ y” + 2y’ + y = 0
Using L.T. on both sides:
L[y"] + 2L[y'] + L[y] = 0
⇒ s2L[y(x)] - sy(0) - y’(0) + 2[SL[y(x)]] - y(0) + L[y] = 0
⇒ (s2 + 2s + 1) L[y(x)] - s(1) - (-1) - 2(1) = 0
[∵ y(0) = 1 & y’(0) = -1]
⇒ (s2 + 2s + 1) L[y(x)] = 1 + s
Applying inverse L.T.