Then the complementary solution is yc= C1e3x+ C2e−x, C1, C2 ∈ R

1. The auxiliary equation is r² − 2r − 3 = 0 ⇒ (r − 3)(r + 1) = 0 ⇒ r = 3 or r = −1. Then the complementary solution is

y_c= C₁e^3x+ C₂e^−x, C₁, C₂ ∈ R.

Take the particular solution y_p = A sin 2x + B cos 2x. Then yp = A sin 2x + B cos 2x,

y⁰_p = 2A cos 2x − 2B sin 2x, y⁰⁰_p = 4A sin 2x − 4B cos 2x,

y⁰⁰_p − 2y_p⁰ − 3yp = (−7A + 4B) sin 2x + (−4A − 7B) cos 2x = cos x.

So we have −7A + 4B = 0 and −4A − 7B = 1 and A = ⁻⁴₆₅ and B = ⁻⁷₆₅. So the general solution is

y = yc+ yp = C1e^3x+ C2e^−x+ −4

65 sin 2x + −7

65 cos 2x, C1, C2 ∈ R.

3. The auxiliary equation is r²+9 = 0 ⇒ (r +3i)(r −3i) = 0 ⇒ r = 3i or r =

−3i. Then the complementary solution is y_c= C₁cos 3x+C₂sin 3x, C₁, C₂ ∈ R. Take the particular solution y_p = Ae^−2x. Then

y_p = Ae^−2x, y_p⁰ = −2Ae^−2x, y_p⁰⁰ = 4Ae^−2x,

y_p⁰⁰+ 9y_p = 13Ae^−2x = e^−2x.

So we have 13A = 1 and A = ₁₃¹. So the general solution is y = y_c+ y_p = C₁cos 3x + C₂sin 3x + 1

13e^−2x, C₁, C₂ ∈ R.

5. The auxiliary equation is r² − 4r + 5 = 0 ⇒ r = 2 ± i. Then the complementary solution is y_c = e^2x(C₁cos x + C₂sin x), C₁, C₂ ∈ R. Take the particular solution yp = Ae^−x. Then

y_p = Ae^−x, y_p⁰ = −Ae^−x, y⁰⁰_p = Ae^−x,

y_p⁰⁰− 4y_p+ 5y_p = 10Ae^−x= e^−x. 1

So we have 10A = 1 and A = ₁₀¹. So the general solution is y = y_c+ y_p = e^2x(C₁cos x + C₂sin x) + 1

10e^−x, C₁, C₂ ∈ R.

13. The auxiliary equation is r²+9 = 0 ⇒ (r+3i)(r−3i) = 0 ⇒ r = 3i or r =

−3i. Then the complementary solution is y_c= C₁cos 3x+C₂sin 3x, C₁, C₂ ∈ R. The trial particular solution is yp = Ae^2x+ (Bx²+ Cx + D) cos x + (Ex²+ F x + G) sin x.

15. The auxiliary equation is r²− 3r + 2 = 0 ⇒ (r − 2)(r − 1) = 0 ⇒ r = 2 or r = 1. Then the complementary solution is y_c = C₁e^2x+ C₂e^x, C₁, C₂ ∈ R.

The trial particular solution is Ae^x+ B sin x + C cos x.

19.(a) The auxiliary equation is 4r²+1 = 0 ⇒ (2r +i)(2r −i) = 0 ⇒ r = ±₂ⁱ. Then the complementary solution is y_c = C₁cos¹₂x + C₂sin¹₂x, C₁, C₂ ∈ R.Take the particular solution y_p = A sin x + B cos x. Then

yp = A sin x + B cos x, y_p⁰ = A cos x − B sin x, y_p⁰⁰= −A sin x − B cos x,

4y⁰⁰_p + yp = −3A sin 2x + −3B cos 2x = cos x.

So we have −3A = 0 and −3B = 1 and A = 0 and B = ⁻¹₃ . So the general solution is

y = y_c+ y_p = C₁cos1

2x + C₂sin1

2x + −1

3 cos x, C₁, C₂ ∈ R.

(b) From (a) we know y_c= C₁e^−2x+ C₂e^−x, C₁, C₂ ∈ R. Set y₁ = cos^x₂ and y2 = sin^x₂. We have y1y⁰₂− y2y⁰₁ = ¹₂cos^{2 x}₂ +¹₂sin^{2 x}₂ = ¹₂. Thus

u⁰₁ = −cos x · sin^x₂

4 · ¹₂ = −1

2 (2 cos² x

2 − 1) sinx 2 u⁰₂ = cos x · cos^x₂

4 · ¹₂ = 1

2(1 − 2 sin² x 2) cosx

2

2

and

u₁ = Z

u⁰₁dx = − cosx 2 −2

3cos³x 2 u₂ =

Z

u⁰₂dx = sinx 2 − 2

3sin³ x 2.

Thus y_p = u₁(x) cos^x₂ + u₂(x) sin^x₂ = ⁻¹₃ cos x and the general solution is

y = y_c+ y_p = C₁cos1

2x + C₂sin1

2x + −1

3 cos x, C₁, C₂ ∈ R.

21. (a) The auxiliary equation is r² − 2r + 1 = 0 ⇒ r = 1. Then the complementary solution is y_c = C₁e^x+C₂xe^x, C₁, C₂ ∈ R. Take the particular solution y_p = Ae^2x. Then

yp = Ae^2x, y_p⁰ = 2Ae^2x, y_p⁰⁰= 4Ae^2x, y_p⁰⁰− 2yp+ yp = Ae^x = e^2x. So we have A = 1 and A = 1. So the general solution is

y = y_c+ y_p = C₁e^x+ C₂xe^x+ e^2x, C₁, C₂ ∈ R.

(b) From (a) we know the complementary solution is y_c= C₁e^x+C₂xe^x, C₁, C₂ ∈ R. Set y1 = e^x, y2 = xe^x. Then y1y₂⁰ − y2y⁰₁ = e^2x and so

u⁰₁ = −xe^x ⇒ u₁(x) = Z

−xe^xdx = −(x − 1)e^x, u⁰₂ = e^x ⇒ u₂(x) =

Z

e^xdx = e^x.

Thus, y_p(x) = (1 − x)e^2x+ xE^2x= e^2x and the general solution is y = y_c+ y_p = C₁e^x+ C₂xe^x+ e^2x, C₁, C₂ ∈ R.

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23. The auxiliary equation is r²+ 1 = 0 ⇒ r = ±i. Then the complementary solution is y_c = C₁cos x + C₂sin x, C₁, C₂ ∈ R. Set y₁ = sin x, y₂ = cos x.

Then y₁y₂⁰ − y₂y⁰₁ = −1 and so u⁰₁ = −sec²x cos x

−1 = sec x ⇒ u₁(x) = Z

sec xdx = ln | sec x + tan x|, (for 0 ≤ x ≤ π 2) u⁰₂ = sec²x sin x

−1 = − sec x tan x ⇒ u₂(x) = Z

− sec x tan xdx = − sec x.

Thus, y_p(x) = ln | sec x + tan x| · sin x − sec x cos x and the general solution is y = y_c+y_p = C₁cos x+C₂sin x+ln | sec x+tan x|·sin x−sec x cos x, C₁, C₂ ∈ R.

25. The auxiliary equation is r²− 3r + 2 = 0 ⇒ (r − 2)(r − 1) = 0 ⇒ r = 2 or r = 1. Then the complementary solution is y_c = C₁e^2x+ C₂e^x, C₁, C₂ ∈ R.

Set y1 = e^x, y2 = e^2x. Then y1y₂⁰ − y2y₁⁰ = e^3x and so u⁰₁ = −e^2x

(1 + e^−x)e^3x = −e^−x

1 + e^−x ⇒ u₁(x) =

Z −e^−x

1 + e^−xdx = ln(1 + e^−x) u⁰₂ = e^x

(1 + e^−x)e^3x = e^x

e^3x+ e^2x ⇒ u2(x) =

Z e^x

e^3x+ e^2xdx = ln(e^x+ 1

e^x ) − e^−x. Thus, y_p(x) = e^xln(1 + e^−x) + e^2x[ln(^ex_e⁺¹x ) − e^−x] and the general solution is y = y_c+y_p = C₁e^2x+C₂e^x+e^xln(1+e^−x)+e^2x[ln(e^x+ 1

e^x )−e^−x], C₁, C₂ ∈ R.

27. The auxiliary equation is r² − 2r + 1 = 0 ⇒ r = 1. Then the comple- mentary solution is yc = C1e^x + C2xe^x, C1, C2 ∈ R. Set y1 = e^x, y2 = xe^x. Then y₁y₂⁰ − y₂y⁰₁ = e^2x and so

u⁰₁ = −xe^{x e}_1+x^x2

e^2x = −x

1 + x² ⇒ u₁(x) =

Z −x

1 + x²dx = −1

2 ln(1 + x²) u⁰₂ = e^{x e}_1+x^x2

e^2x = 1

1 + x² ⇒ u₂(x) =

Z 1

1 + x²dx = tan⁻¹x.

Thus, y_p(x) = e^{x −1}₂ ln(1 + x²) + xe^xtan⁻¹x and the general solution is y = y_c+y_p = C₁e^x+C₂xe^x+e^xln(1+e^−x)+e^x−1

2 ln(1+x²)+xe^xtan⁻¹x, C₁, C₂ ∈ R.

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