Homework 3.4 - Second-Order Linear Homogenous Differential Equations

Determine whether the following equations are second-order linear differential equations:

a) y′′=−4x(y′)2y^{\prime\prime} = -4x(y^\prime)²

  • No

b) 8 \frac {d²y}{dx²} - 6 \frac {dy}{dx}  + 11y = 0

  • Yes

c) d2ydx2=xsin⁡(y)\frac {d²y}{dx²} = x \sin (y)

  • No

d) y′′=3yy′y^{\prime\prime} = 3yy^\prime

  • No

e) 5y′′−12y=7x2+9x5y^{\prime\prime} - 12y = 7x² + 9x

  • Yes

Find the general solution to the differential equation

y′′−9y′+20y=0y^{\prime\prime} - 9y^\prime + 20y = 0

a=1a = 1

b=−9b = -9

c=20c = 20

(y−4)(y−5)(y-4)(y-5)

r=4,5r = 4, 5

General solution example: y(x)=C1er1x+C2er2xy(x) = C_1e^{r_1x} + C_2 e^{r_2 x}

General solution: y(x)=C1e4x+C2e5xy(x) = C_1e^{4x} + C_2e^{5x}

Find the general solution to the differential equation

y′′+6y′+34y=0y^{\prime\prime} + 6y^{\prime} + 34 y = 0

a=1a = 1

b=6b = 6

c=34c = 34

b2−4ac⇒62−4(1)(34)b²-4ac \Rightarrow 6²-4(1)(34)

36 - 136 < 0

Imaginary roots

−6±−1002\frac {-6 \pm \sqrt {-100}}{2}

  • −6−10i2\frac {-6 - 10i}{2}

  • −6+10i2\frac {-6 + 10i}{2}

y(x)=C1e−6−10i2x+C2e−6+10i2xy(x) = C_1e^{\frac {-6-10i}{2}x} + C_2e^{\frac {-6+10i}{2}x}

Find the general solution to the differential equation

16d2ydx2+56dydx+49y=016\frac{d²y}{dx²}+56\frac{dy}{dx}+49y=0

a=16a=16

b=56b=56

c=49c=49

b2−4ac⇒562−4(16)(49)b²-4ac\Rightarrow 56² - 4(16)(49)

⇒0\Rightarrow 0

One real solution

−56± 036\frac {-56\pm\ 0}{36} ⇒−1.75\Rightarrow -1.75

y(x)=C1e−1.75x+C2xe−1.75xy(x) = C_1 e^{-1.75x} + C_2 xe^{-1.75x}

Solve the initial-value problem

3y′′+5y′+2y=03y^{\prime\prime} + 5y^{\prime} + 2y = 0

y(0)=5y(0) = 5

y′(0)=−4y^{\prime}(0) = -4

a=3a=3

b=5b=5

c=2c=2

b2−4ac⇒b²-4ac\Rightarrow 25−24=125-24 = 1
−b±b2−4ac2a\frac {-b\pm\sqrt{b²-4ac}}{2a}

−5±16\frac {-5\pm 1}{6} ⇒−46\Rightarrow \frac{-4}6 and −1-1

 

y(x)=C1e−23x+C2e−1xy(x) = C_1e^{-\frac {2}{3}x} + C_2e^{-1x}

⇒y(0)=C1e−23⋅0+C2e−0\Rightarrow y(0) = C_1 e^{-\frac 23 \cdot 0} + C_2 e^{-0}

⇒5=C1+C2\Rightarrow 5 = C_1 + C_2

Derivative of y(x)y(x) is −23C1e−23x−1C2e−x-\frac 23 C_1 e^{-\frac 23 x} -1 C_2e^{-x}

⇒y′(x)=−23C1e0−1C2e0\Rightarrow y^\prime (x) = -\frac 23 C_1e^0 - 1C_2e^0

⇒−4=−23C1−C2\Rightarrow -4 = -\frac 23 C_1 -C_2

Solve the system of equations:

5=C1+C25 = C_1+C_2
−4=−23C1−C2-4 = -\frac 23 C_1 - C_2

5+(−4)=C1+C2−23C1−C25 + (-4) = C_1 + C_2 -\frac 23 C_1 - C_2

1=C1−23C11 = C_1 - \frac 23 C_1

1=13C11=\frac 13 C_1

3=C13 = C_1

5=3+C25 = 3 + C_2

2=C22 = C_2

y(x)=C1e−23x+C2e−x⇒y(x) = C_1e^{-\frac 23 x} + C_2e^{-x} \Rightarrow y(x)=3e−23x+2e−xy(x) = 3e^{-\frac 23 x} + 2e^{-x}

Solve the initial-value problem

y′′−6y′+10y=0y^{\prime\prime} - 6y^{\prime} + 10y = 0

y(0)=2y(0) = 2

y′(0)=4y^\prime(0) = 4

a=1a = 1

b=−6b = -6

c=10c = 10

b2−4ac⇒36−4(1)(10)=−4b² - 4ac \Rightarrow 36 - 4(1)(10) = -4

−b±b2−4ac2a⇒6±−42\frac {-b \pm \sqrt {b²-4ac}}{2a} \Rightarrow \frac {6\pm \sqrt{-4}}{2}

⇒6+2i2\Rightarrow \frac {6+2i}{2} and 6−2i2⇒3+i\frac {6-2i}{2} \Rightarrow 3+i and 3−i3-i

α=3\alpha = 3

β=1\beta = 1 

y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin (\beta x))

y^\prime(x) = 

Solve the initial-value problem

y′′+4y=0y^{\prime\prime} + 4y = 0

y(π2)=0y(\frac {\pi}2) = 0

y′(π2)=3y^\prime (\frac \pi 2) = 3

Solve the boundary-value problem

y′′+18y′+81y=0y^{\prime\prime} + 18y^\prime + 81y =0

y(0)=0y(0) = 0

y(1)=3y(1) = 3

Simplify the complex number completely

(−8−5i)(2+9i)(-8-5i)(2+9i)

Simplify the complex number completely

4+i3+2i\frac {4+i}{3+2i}