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3 years ago

1.03 The constitution

Advanced Placement (AP)

1 answer:

iris [78.8K]3 years ago

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Answer:

The Constitution of the United States, approved in 1787 by the Constitutional Convention of Philadelphia, is the supreme law of the United States, which defines the political organization of the nation, the structure of the federal government and, based on its amendments, the individual rights of its citizens.

This Constitution originally consisted of 7 articles of a programmatic nature, which define the roles and structure of the government. In addition, 27 amendments were subsequently added, guaranteeing certain civil rights to American citizens. An important role is held by the first 10 amendments, which form a normative body called the Bill of Rights, with the fundamental civil rights of Americans.

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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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