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MTHE 225
HOMEWORK SET 11
1. Solve the differential equation: 2y 00 + 3y 0 − 2y = 14x2 − 4x − 11.
2. Solve the differential equation: y 00 − y 0 − 2y = x + e2x .
3. Solve the differential equation: y 00 + y = 3 sin x.
4. Solve the differential equation: 4y 00 + 4y 0 + y = 3xex .
1
SOLUTION
1.
(Non-homogeneous Linear DE with Constant Coefficients)
• For yc :
Auxiliary equation: 2m2 + 3m − 2 = 0 ⇒ (2m − 1)(m + 2) = 0 ⇒ m =
1
⇒ yc = C1 e 2 x + C2 e−2x .
1
, −2
2
• For yp :
Let yp = A2 x2 + A1 x + A0 .
⇒ yp0 = 2A2 x + A1
⇒ yp00 = 2A2 .
Now plug these in the given DE.
2(2A2 ) + 3(2A2 x + A1 ) − 2(A2 x2 + A1 x + A0 ) = 14x2 − 4x − 11
⇒ (−2A2 )x2 + (6A2 − 2A1 )x + (4A2 + 3A1 − 2A0 ) = 14x2 − 4x − 11
Now compare the coefficients.
−2A2 = 14 ⇒ A2 = −7.
6A2 − 2A1 = −4 ⇒ −42 − 2A1 = −4 ⇒ A1 = −19.
4A2 + 3A1 − 2A0 = −11 ⇒ −28 − 57 − 2A0 = −11 ⇒ A0 = −37.
Therefore, yp = −7x2 − 19x − 37.
1
Hence the general solution of the DE is y(x) = yc + yp = C1 e 2 x + C2 e−2x − 7x2 − 19x − 37.
2.
(Non-homogeneous Linear DE with Constant Coefficients)
• For yc :
Auxiliary equation: m2 − m − 2 = 0 ⇒ (m − 2)(m + 1) = 0 ⇒ m = −1, 2
⇒ yc = C1 e−x + C2 e2x .
• For yp :
not Be2x
z }| {
Let yp = (A1 x + A0 ) + Bxe2x
⇒ yp0 = A1 + B(e2x + 2xe2x ) = A1 + B(1 + 2x)e2x
⇒ yp00 = B(2e2x + (1 + 2x)2e2x ) = B(4 + 4x)e2x .
Now plug these in the given DE.
(B(4 + 4x)e2x ) − (A1 + B(1 + 2x)e2x ) − 2((A1 x + A0 ) + Bxe2x ) = x + e2x
⇒ (−2A1 )x + (−A1 − 2A0 ) + B(4 + 4x − 1 − 2x − 2x)e2x = x + e2x
2
⇒ (−2A1 )x + (−A1 − 2A0 ) + B(3)e2x = x + e2x
Now compare the coefficients.
1
−2A1 = 1 ⇒ A1 = − .
2
1
1
−A1 − 2A0 = 0 ⇒ − 2A0 = 0 ⇒ A0 = .
2
4
1
3B = 1 ⇒ B = .
3
1
1 1
Therefore, yp = − x + + xe2x .
2
4 3
1
1 1
Hence the general solution of the DE is y(x) = yc + yp = C1 e−x + C2 e2x − x + + xe2x .
2
4 3
3.
(Non-homogeneous Linear DE with Constant Coefficients)
• For yc :
√
Auxiliary equation: m2 +1 = 0 ⇒ m2 −( −1)2 = 0 ⇒ m2 −i2 = 0 ⇒ (m−i)(m+i) = 0
⇒ m = ±i. (Recall any complex/imaginary root(s) of a polynomial comes in a pair of
form α ± βi)
⇒ yc = C1 eαx cos (βx) + C2 eαx sin (βx) = C1 e0x cos (1x) + C2 e0x sin (1x) = C1 cos x +
C2 sin x.
• For yp :
not (A cos x+B sin x)
z
}|
{
Let yp = x(A cos x + B sin x)
⇒ yp0 = (A cos x + B sin x) + x(−A sin x + B cos x)
⇒ yp00 = (−A sin x + B cos x) + (−A sin x + B cos x) + x(−A cos x − B sin x)
⇒ yp00 = 2(−A sin x + B cos x) + x(−A cos x − B sin x)
Now plug these in the given DE.
2(−A sin x + B cos x) + x(−A cos x − B sin x) + x(A cos x + B sin x) = 3 sin x
⇒ −2A sin x − 2B cos x = 3 sin x
Now compare the coefficients.
3
−2A = 3 ⇒ A = −
2
−2B = 0 ⇒ B = 0
3
Therefore, yp = − x cos x
2
3
Hence the general solution of the DE is y(x) = yc + yp = C1 cos x + C2 sin x − x cos x.
2
4.
(Non-homogeneous Linear DE with Constant Coefficients)
3
• For yc :
1 1
Auxiliary equation: 4m2 + 4m + 1 = 0 ⇒ (2m + 1)2 = 0 ⇒ m = − , − .
2 2
1
1
⇒ yc = C1 e− 2 x + C2 xe− 2 x .
• For yp :
Let yp = (Ax + B)ex
⇒ yp0 = Aex + (Ax + B)ex = (Ax + A + B)ex
⇒ yp00 = Aex + (Ax + A + B)ex = (Ax + 2A + B)ex
Now plug these in the given DE.
4(Ax + 2A + B)ex + 4(Ax + A + B)ex + (Ax + B)ex = 3xex
⇒ (9Ax + 12A + 9B)ex = 3xex
⇒ 9Ax + 12A + 9B = 3x
Now compare the coefficients.
1
9A = 3 ⇒ A =
3
1
4
12A + 9B = 0 ⇒ 12 · + 9B = 0 ⇒ B = −
3
9
1
4 x
Therefore, yp =
x−
e
3
9
Hence the general solution of the DE is y(x) = yc +yp = C1 e
4
− 21 x
+ C2 xe
− 12 x
+
1
4
x−
3
9
ex .