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4 years ago

What percentage of the Department of Transportation's budget is for winter road maintance?

1 answer:

5 0

The percentage of the Department of Transportation's budget is for winter road maintenance is 20%..State DOTs (Department of Transportation) in snowy regions spend more of their budget on clearing roads during the winter season.The high percentage of the budget is necessary to reduce accidents due to loss of friction between roads and automobile tires, and to repair damage to roads. Even with this high spending rate, many accidents occur in the winter because of poor visibility and loss of traction due to slipping of automobiles.

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4 years ago

Solve the following differential equation with initial conditions: y''=e^-2t+10e^4t ; y(0)=1, y'(0)=0

skad [1K]

Answer:

Option A. $y = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

Explanation:

This is a second order DE, so we'll need to integrate twice, applying initial conditions as we go. At a couple points, we'll need to apply u-substitution.

Round 1:

To solve the differential equation, write it as differentials, move the differential, and integrate both sides:

$y''=e^{-2t}+10e^{4t}$

$\frac{dy'}{dt}=e^{-2t}+10e^{4t}$

$dy'=[e^{-2t}+10e^{4t}]dt$

$\int dy'=\int [e^{-2t}+10e^{4t}]dt$

Applying various properties of integration:

$\int dy'=\int e^{-2t} dt + \int 10e^{4t}dt\\\int dy'=\int e^{-2t} dt + 10\int e^{4t}dt$

Prepare for integration by u-substitution

$\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt$ , letting $u_1=-2t$ and $u_2=4t$

Find dt in terms of $u_1 \text{ and } u_2$

$u_1=-2t\\du_1=-2dt\\-\frac{1}{2}du_1=dt$ $u_2=4t\\du_2=4dt\\\frac{1}{4}du_2=dt$

$\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt\\\int dy'=\int e^{u_1} (-\frac{1}{2} du_1) + 10\int e^{u_2} (\frac{1}{4} du_2)\\\int dy'=-\frac{1}{2} \int e^{u_1} (du_1) + 10 *\frac{1}{4} \int e^{u_2} (du_2)$

Using the Exponential rule (don't forget your constant of integration):

$y'=-\frac{1}{2} e^{u_1} + 10 *\frac{1}{4}e^{u_2} +C_1$

Back substituting for $u_1 \text{ and } u_2$ :

$y'=-\frac{1}{2} e^{(-2t)} + 10 *\frac{1}{4}e^{(4t)} +C_1\\y'=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} +C_1\\$

Finding the constant of integration

Given initial condition $y'(0)=0$

$y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} +C_1\\0=y'(0)=-\frac{1}{2} e^{-2(0)} + \frac{5}{2}e^{4(0)} +C_1\\0=-\frac{1}{2} (1) + \frac{5}{2}(1) +C_1\\-2=C_1\\$

The first derivative with the initial condition applied: $y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\$

Round 2:

Integrate again:

$y' =-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\\frac{dy}{dt} =-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\dy =[-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2]dt\\\int dy =\int [-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2]dt\\\int dy =\int -\frac{1}{2} e^{-2t} dt + \int \frac{5}{2}e^{4t} dt - \int 2 dt\\\int dy = -\frac{1}{2} \int e^{-2t} dt + \frac{5}{2} \int e^{4t} dt - 2 \int dt\\$

$y = -\frac{1}{2} * -\frac{1}{2} e^{-2t} + \frac{5}{2} * \frac{1}{4} e^{4t} - 2 t + C_2\\y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + C_2$

Finding the constant of integration :

Given initial condition $y(0)=1$

$1=y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + C_2\\1 = \frac{1}{4} (1) + \frac{5}{8} (1) - (0) + C_2\\1 = \frac{7}{8} + C_2\\\frac{1}{8}=C_2$

So, $y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

Checking the solution

$y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}$

This matches our initial conditions here $y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + \frac{1}{8} = 1$

Going back to the function, differentiate:

$y' = [\frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}]'\\y' = [\frac{1}{4} e^{-2t}]' + [\frac{5}{8} e^{4t}]' - [2 t]' + [\frac{1}{8}]'\\y' = \frac{1}{4} [e^{-2t}]' + \frac{5}{8} [e^{4t}]' - 2 [t]' + [\frac{1}{8}]'$

Apply Exponential rule and chain rule, then power rule

$y' = \frac{1}{4} e^{-2t}[-2t]' + \frac{5}{8} e^{4t}[4t]' - 2 [t]' + [\frac{1}{8}]'\\y' = \frac{1}{4} e^{-2t}(-2) + \frac{5}{8} e^{4t}(4) - 2 (1) + (0)\\y' = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2$

This matches our first order step and the initial conditions there.

$y'(0) = -\frac{1}{2} e^{-2(0)} + \frac{5}{2} e^{4(0)} - 2=0$

Going back to the function y', differentiate:

$y' = -\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2\\y'' = [-\frac{1}{2} e^{-2t} + \frac{5}{2} e^{4t} - 2]'\\y'' = [-\frac{1}{2} e^{-2t}]' + [\frac{5}{2} e^{4t}]' - [2]'\\y'' = -\frac{1}{2} [e^{-2t}]' + \frac{5}{2} [e^{4t}]' - [2]'$

Applying the Exponential rule and chain rule, then power rule

$y'' = -\frac{1}{2} e^{-2t}[-2t]' + \frac{5}{2} e^{4t}[4t]' - [2]'\\y'' = -\frac{1}{2} e^{-2t}(-2) + \frac{5}{2} e^{4t}(4) - (0)\\y'' = e^{-2t} + 10 e^{4t}$

So our proposed solution is a solution to the differential equation, and satisfies the initial conditions given.

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2 years ago

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