Differential Equations Answers

Solve the initial value problem y"-3y'-2y=0; y(0)=1, y(3)=0

(D^3 + 3D^2-D-3)y = e^2x+4

Reducible to homogeneous differential equations or by simple substitution.

1. (2𝑥 − 3𝑦 − 4)𝑑𝑥 − (3𝑥 − 4𝑦 − 2)𝑑𝑦 = 0

2. (2𝑥 − 𝑦 − 3)𝑑𝑥 − (𝑥 + 4𝑦 + 3)𝑑𝑦 = 0

3. (𝑥 − 𝑦 − 6)𝑑𝑦 = (𝑥 − 𝑦 + 2)𝑑𝑥

4. (𝑥 − 2𝑦 + 4)𝑑𝑥 + (2𝑥 − 4𝑦 − 1)𝑑𝑦 = 0

5. (𝑥 + 4𝑦 + 3)𝑑𝑥 = (2𝑥 − 𝑦 − 3)𝑑𝑦

B. Exact differential equations (include checking for exactness).

1. (𝑤3 + 𝑤𝑧 2 − 𝑧)𝑑𝑤 + (𝑧 2 + 𝑤2 𝑧 − 𝑤)𝑑𝑧 = 0

2. (cos 2𝑦 − 3𝑥 2𝑦 2 )𝑑𝑥 + (cos 2𝑦 − 2𝑥 sin 2𝑦 − 2𝑥 3𝑦)𝑑𝑦 = 0

1.  y'' + 4y' + 7y = 0         ans, y= e^-2x(c₁cos"sqr3" x+c₂sin"sqr3" x)

1.  y'''-4y'=0                   ans, y=c₁+c₂e^2x+c₃e^-2x
2. y'' + 9y' + 5y = 0 

1. y'''+3y''+4y'-8y=0    ans, y= c₁e^x+e^-2x(c₂cos2x+c₃sin2x)
2.  y⁽⁴⁾-4y''=0                   ans, y=c₁+c₂x+c₃e^2x+c₄e^-2x

1.  y⁽⁴⁾+8y''+16y=0       ans, y=(c₁+c₂x)cos2x+(c₃+c₄x)sin2x
2. y⁽⁴⁾+4y''+4y=0          ans, y=(c₁+c₂x)cos "sqr2" x+(c₃+c₄x)sin"sqr2" x

1.    y''' - 9y''+ 27y'-27y=0        ans, y=c₁e^3x+c₂xe^3x+c₃x²e^3x

2.    y''' + 2y''- 13y' + 10y =0    ans, y= c₁e^-5x+c₂e^2x+c₃e^x

1. y'' - 2y' = 0                ans, y=c₁+c₂e^2x
2. 4y'' - 4y' + y = 0       ans, y=c₁e^1/2x+c₂xe^1/2x