Part 3: Differential Equations

For each differential equation below, solve the initial value problem

Solve the IVP: y′=3x2y^\prime = 3x²,    y(0)=0y(0) = 0

∫y′dy=∫3x2dx\int y^\prime dy = \int 3x² dx

  • y=x3+Cy = x³ + C

0=03+C0 = 0³ + C

  • C=0C = 0

y=x3y= x³

Check whether y=Ce5xy=Ce^5x is a solution to y′=5yy^\prime = 5y

y′=5Ce5xy^\prime = 5Ce^5x

  • This is a solution

Solve the IVP y′=xyy^\prime = xy,    y(0)=1y(0) = 1

dydx=xy\frac {dy}{dx} = xy

  • dyy=dxx\frac {dy}{y} = dx x

∫1ydy=∫xdx\int \frac 1y dy = \int x dx

  • ln⁡∣y∣=x22+C\ln |y| = \frac {x²}2 + C

eln⁡∣y∣=ex22Ce^{\ln |y|} = e^{\frac {x²}{2}}C

  • y=Cex22y = Ce^{\frac {x²}{2}}

1=Ce01 = Ce^{0}

  • 1=C11 = C1

  • C=1C = 1

y=ex22y=e^{\frac {x²}2}

Population growth: dPdt=kP\frac {dP}{dt} = kP,    P(0)=10P(0) = 10

dPP=kdt\frac {dP}{P} = k dt

  • ∫1PdP=∫kdt\int \frac 1{P} dP = \int k dt

  • ln⁡∣P∣=kt+C\ln |P| = kt + C

eln⁡∣P∣=Cekte^{\ln |P|} = Ce^{kt}

  • P=CektP = Ce^{kt}

10=Cek(0)10 = Ce^{k(0)}

  • 10=C10=C

P(t)=10ektP(t) = 10e^{kt}

Newton’s Cooling Law: dTdt=−k(T−20)\frac {dT}{dt} = -k(T-20),    T(0)=80T(0) = 80

dTT−20=−kdt\frac {dT}{T-20} = -k dt

  • ∫1T−20dT=∫−kdt\int \frac 1{T-20} dT = \int -k dt

  • ln⁡∣T−20∣=−kt+C\ln |T-20| = -kt + C

eln⁡∣T−20∣=Ce−kte^{\ln|T-20|} = Ce^{-kt}

  • T−20=Ce−ktT-20 = Ce^{-kt}

  • T(t)=Ce−kt+20T(t) = Ce^{-kt} + 20

80=Ce−k(0)+2080 = Ce^{-k(0)} + 20

  • 60=C60 = C

T(t)=60e−kt+20T(t) = 60e^{-kt} + 20

Solve the IVP: y′′−4y=0y^{\prime\prime} - 4y = 0,    y(0)=1y(0)= 1,    y′(0)=0y^\prime(0) = 0

Characteristic: r2−4=0⇒r=±2r²-4=0 \Rightarrow r = \pm 2

y=C1e2x+C2e−2xy = C_1e^{2x} + C_2e^{-2x}

y(0)=C1+C2=1y(0) = C_1 + C_2 = 1

  • C1=C2=12C_1 = C_2 = \frac 12

y′(0)=2C1−2C2=0y^\prime (0) =2C_1 - 2C_2 = 0

y=12e2x+12e−2xy = \frac 12e^{2x} + \frac 12 e^{-2x}

Solve the IVP: y′′−6y′+9y=0y^{\prime\prime} - 6y^\prime + 9y = 0,    y(0)=2y(0) = 2,    y′(0)=0y^\prime (0)= 0

a=1a = 1

b=−6b = -6

c=9c=9

b2−4ac⇒36−36=0b² - 4ac \Rightarrow 36 - 36 = 0

  • One root

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

  • 6±02\frac {6 \pm 0}{2}

  • 33

y=C1e3x+C2xe3xy = C_1e^{3x} + C_2xe^{3x}

y(0)=C1=2y(0) = C_1 = 2

y′(0)=6e3(0)+(C2e3(0))+(C2(0)3e(0))=0y^\prime (0) = 6e^{3(0)} + (C_2e^{3(0)}) + (C_2 (0) 3e^{(0)}) = 0

  • 6+C2=06 + C_2 = 0

  • C2=−6C_2 = -6

y(x)=2e3x−6xe3xy(x) = 2e^{3x} -6xe^{3x}

Solve the IVP: y′′+5y′+6y=0y^{\prime\prime} + 5y^\prime + 6y = 0,    y(0)=1y(0) = 1,    y′(0)=1y^\prime(0) = 1

a=1a = 1

b=5b = 5

c=6c = 6

b2−4ac⇒25−24=1b² - 4ac \Rightarrow 25 - 24 = 1

  • Two real roots

−b±b2−4ac2a⇒−5±12\frac {-b \pm \sqrt {b²-4ac}}{2a} \Rightarrow \frac {-5 \pm 1}{2}

  • Roots: −2,-2, −3-3

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

y(0)=C1+C2=1y(0) = C_1 + C_2 = 1

y′(0)=−2C1+−3C2=1y^{\prime} (0)= -2C_1 + -3C_2 = 1

C1+C2=−2C1−3C2=1C_1 + C_2 = -2C_1 -3C_2 = 1

  • 3C1=−4C23C_1 = -4C_2

  • −34C1=C2-\frac 34 C_1 = C_2

  • C1−34C1=1C_1 -\frac 34C_1 = 1

  • C1(1−34)=1C_1(1 - \frac 34) = 1

  • C1(14)=1C_1(\frac 14) = 1

  • C1=4C_1 = 4

  • 4+C2=14 + C_2 = 1

  • C2=−3C_2 = -3

y(x)=4e−2x−3e−3xy(x) = 4e^{-2x} -3e^{-3x}

Solve the IVP: y′′−7y′+10y=0y^{\prime\prime} -7y^\prime +10y = 0,    y(0)=1y(0) = 1,    y′(0)=0y^\prime(0) = 0

a=1a = 1

b=−7b= -7

c=10c=10

b2−4ac⇒49−40=9b²-4ac \Rightarrow 49 - 40 = 9

  • Two real roots

7±32\frac {7 \pm 3} {2}

  • Roots: 5,5, 22

y(x)=C1e5x+C2e2xy(x) = C_1e^{5x} + C_2e^{2x}

y(0)=C1+C2=1y(0) = C_1 + C_2 = 1

y′(0)=5C1+2C2=0y^\prime (0) = 5C_1 + 2C_2 = 0

C1+C2=5C1+2C2+1C_1 + C_2 = 5C_1 + 2C_2 + 1

  • −4C1=C2+1-4C_1 = C_2 + 1

  • C2=−4C1−1C_2 = -4C_1 - 1

  • C1+(−4C1−1)=1C_1 + (-4C_1 - 1) = 1

  • C1−4C1−1=1C_1 -4C_1 - 1 = 1

  • C1(1−4)=2C_1(1 - 4) = 2

  • C1=−23C_1 = -\frac 23

  • C2=53C_2 = \frac 53

y(x)=−23e5x+53e2xy(x) = -\frac 23 e^{5x} + \frac 53 e^{2x}

Solve the IVP: y′′+2y′+y=0y^{\prime\prime} + 2y^{\prime} + y = 0,    y(0)=1y(0) = 1,    y′(0)=0y^{\prime}(0) = 0

a=1a=1

b=2b=2

c=1c=1

b2−4ac=0b² - 4ac = 0

  • One real root

Root(s) = −1-1

y(x)=C1e−x+C2xe−xy(x) = C_1e^{-x} + C_2 x e^{-x}

y(0)=C1=1y(0) = C_1 = 1

y′(t)=−C1e−x+C2(e−x)+C2x(−e−x)y^{\prime} (t) = -C_1e^{-x} + C_2(e^{-x}) + C_2x(-e^{-x})

  • y′(0)=−C1+C2=0y^{\prime}(0) = -C_1 + C_2 = 0

  • −1+C2=0-1 + C_2 = 0

  • C2=1C_2 = 1

y(x)=e−x+xe−xy(x) = e^{-x} + xe^{-x}

Solve the IVP: y′′−9y=0y^{\prime\prime} - 9y = 0    y(0)=2y(0) = 2,    y′(0)=0y^\prime(0) = 0

a=1a = 1

b=0b = 0

c=−9c = -9

b2−4ac=36b² - 4ac = 36

±62\frac {\pm 6}{2}

  • Roots: −3,3-3, 3

y(x)=C1e−3x+C2e3xy(x) = C_1e^{-3x} + C_2e^{3x}

y(0)=C1+C2=2y(0) = C_1 + C_2 = 2

y′(t)=−3C1e−3x+3C2e3xy^{\prime}(t) = -3C_1e^{-3x} + 3C_2e^{3x}

  • y′(0)=−3C1+3C2=0y^\prime (0) = -3C_1 + 3C_2 = 0

  • C1=C2=1C_1 = C_2 = 1

y(x)=e−3x+e3xy(x) = e^{-3x} + e^{3x}

Solve the IVP: y′′+16y=0y^{\prime\prime} + 16y = 0,    y(0)=0y(0) = 0,    y′(0)=4y^\prime(0) = 4

a=1a = 1

b=0b = 0

c=16c=16

±8i2\frac {\pm 8i}{2}

r=α±iβr = \alpha \pm i\beta

  • r1=0−4i=−4ir_1 = 0 - 4i = -4i

  • r2=0+4i=4ir_2 = 0 + 4i = 4i

y(x)=C1eαx(cos⁡(βx)+C2sin⁡(βx)y(x) = C_1e^{\alpha x}(\cos(\beta x) + C_2 \sin (\beta x)

  • y(x)=C1cos⁡(−4x)+C2sin⁡(4x)y(x) = C_1\cos(-4x) + C_2 \sin (4 x)

y(0)=C1=0y(0) = C_1 = 0

y′(x)=4C2cos⁡(4x)y^\prime (x) = 4 C_2\cos(4 x)

  • y′(0)=4C2cos⁡(0)=4y^\prime(0) = 4C_2 \cos(0) = 4

  • C2=1C_2 = 1

y(x)=sin⁡(4x)y(x) = \sin(4x)