posted 10 months ago

\dot{x}_{t}=\frac{x_{t}}{x_{t}^{2}+1}

\dot{x}_{t}=x_{t} y_{t}

\dot{y}_{t}=-2 x_{t}-4 y_{t}+4

\dot{x}_{t}=3 x_{t}-13 y_{t}

\dot{y}_{t}=5 x_{t}+y_{t}

x_{t+1}=2-x_{t} .

posted 10 months ago

\bar{x}_{t}=5 x_{t}-t^{2}

x_{1}=1

\dot{x}_{t}=\frac{x_{t}}{t}

x_{1}=1

2 \ddot{x}_{t}+3 \dot{x}_{t}-2 x_{t}=0,

x_{0}=3

\dot{x}_{\mathbf{0}}=-\mathbf{1}

\dot{x}_{t}=7 x_{t}+y_{t}

\dot{y}_{t}=-4 x_{t}+3 y_{t}

x_{0}=2

y_{0}=-5

\dot{x}_{t}=6 x_{t}-3 y_{t},

\dot{y}_{t}=-2 x_{t}+y_{t}

\boldsymbol{x}_{t+1}=\boldsymbol{y}_{t}

y_{t+1}=-x_{t}+5 y_{t}

posted 10 months ago

\frac{d T}{d t}=\ln \left(\frac{1}{2}\right)(T-16), \quad T(0)=70

\text { the ambient temperature } T_{m} \text { is } 16 .

True--False

posted 10 months ago

y^{\prime \prime}-64 y=0, y(0)=1, y^{\prime}(0)=8 .

True--False

posted 10 months ago

y^{\prime}-20=y e^{2 x}, y(1)=5

Using the Euler's method we have

y_{1}=5+h\left(20+5 e^{2}\right)

posted 10 months ago

y^{\prime \prime}+16 y=0, y(0)=0, y^{\prime}(\pi)=4

\sin 4 x

-\sin 4 x

\text { None of the others }

\cos 4 x

\cos 4 x+\sin 4 x

posted 10 months ago

P(t)=P_{0} e^{0.25 t}

If P(5) = 200, what is the initial population?

Select one:

Po - 59

None of the others

Po - 61

Po - 60

Po 57

posted 10 months ago

A) (8 pts) Use Euler's method to obtain an approximation of y(0.5) usingh = 0.25 for the given IVP (Use four-decimal approximation)

B) (12 pts) Use Euler's method to obtain an approximation of y(0.5) usingh = 0.1 for the given IVP (Use four-decimal approximation)

A bacteria culture initially has 100 number of bacteria and doubles in size in 2hours. Assume that the rate of increase of the culture is proportional to the size.

Write the initial value problem for the bacteria culture and solve it

ts) How long will it take for the size to triple?

Verify that y(x) = c1 cos(6x) + c2 sin(6x) is a solution of

y" + 36y = 0

s) Either solve the boundary value problem

y^{\prime \prime}+36 y=0, y(0)=0, y\left(\frac{2 \pi}{6}\right)=1

or else show that it has no solution

posted 10 months ago

y^{\prime}=x+2 y, y(1)=2

numerically using Euler's method for y(1.6) using h = 0.3 is

-5.99

3.5

-3.5

5.99

None of the others

posted 10 months ago

y^{\prime \prime}-5 y^{\prime}+2 y=0

\text { with } y(0)=1 \text { and } y^{\prime}(0)=5 \text { is }

a second order initial value problem

a fourth order initial value problem

a third order initial value problem

None of the others

A boundary value problem