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

Solve The Following Differential Equation With Initial Conditions: Y''=e^-2t+10e^4t ; Y(0)=1, Y'(0)=0

Answers

Answer 1

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.

Answer 2

Answer:

\(\textsf{A.} \quad y=\dfrac{1}{4}e^{-2t}+\dfrac{5}{8}e^{4t}-2t+\dfrac{1}{8}\)

Explanation:

Given:

\(\begin{cases}y''=e^{-2t}+10e^{4t}\\y\:\!(0)=1\\y'(0)=0\end{cases}\)

To find y', integrate y'' and use the condition y'(0) = 0.

\(\begin{aligned}\displaystyle y'=\int y''&=\int e^{-2t}+10e^{4t}\; dt\\\\&=\int e^{-2t}\; dx+10 \int e^{4t}\; dt\\\\&=-\dfrac{1}{2}e^{-2t}+10\cdot \dfrac{1}{4}e^{4t}+C\\\\&=-\dfrac{1}{2}e^{-2t}+\dfrac{5}{2}e^{4t}+C\end{aligned}\)

To find the value of the constant of integration, C, substitute the given condition y'(0) = 0:

\(\begin{aligned}\displaystyle y'(0)&=0\\\\-\dfrac{1}{2}e^{-2(0)}+\dfrac{5}{2}e^{4(0)}+C&=0\\\\-\dfrac{1}{2}(1)+\dfrac{5}{2}(1)+C&=0\\\\-\dfrac{1}{2}+\dfrac{5}{2}+C&=0\\\\2+C&=0\\\\C&=-2\end{aligned}\)

Therefore, the equation for y' is:

\(y'=-\dfrac{1}{2}e^{-2t}+\dfrac{5}{2}e^{4t}-2\)

To find y, integrate y' and use the condition y(0) = 1:

\(\begin{aligned}\displaystyle y=\int y'&=\int -\dfrac{1}{2}e^{-2t}+\dfrac{5}{2}e^{4t}-2\;dt\\\\&=-\dfrac{1}{2}\int e^{-2t}\;dt+\dfrac{5}{2}\int e^{4t}\;dt-\int 2\;dt\\\\&=-\dfrac{1}{2} \cdot -\dfrac{1}{2}e^{-2t}+\dfrac{5}{2}\cdot \dfrac{1}{4}e^{4t}-2t+C\\\\&=\dfrac{1}{4}e^{-2t}+\dfrac{5}{8}e^{4t}-2t+C\end{aligned}\)

To find the value of the constant of integration, C, substitute the given condition y(0) = 1:

\(\begin{aligned}\displaystyle y(0)&=1\\\\\dfrac{1}{4}e^{-2(0)}+\dfrac{5}{8}e^{4(0)}-2(0)+C&=1\\\\\dfrac{1}{4}(1)+\dfrac{5}{8}(1)-0+C&=1\\\\\dfrac{1}{4}+\dfrac{5}{8}+C&=1\\\\\dfrac{7}{8}+C&=1\\\\C&=\dfrac{1}{8}\end{aligned}\)

Therefore the equation for y is:

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

\(\hrulefill\)

Integration rules used:

\(\boxed{\begin{minipage}{5.1 cm}\underline{Terms multiplied by a constant}\\\\$\displaystyle \int a\;\!f(x)\:\text{d}x=a \int f(x)\:\text{d}x$\\\end{minipage}}\)

\(\boxed{\begin{minipage}{5.1 cm}\underline{Integrating $e^{ax}$}\\\\$\displaystyle \int e^{ax}\:\text{d}x=\dfrac{1}{ax}e^{ax}+\text{C}$\\\end{minipage}}\)

\(\boxed{\begin{minipage}{5.1 cm}\underline{Integrating a constant}\\\\$\displaystyle \int n\:\text{d}x=nx+\text{C}$\\\\(where $n$ is any constant value) \end{minipage}}\)


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on earth.

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plant cover provided food and protection for organisms riparian zone-buffers water ... The Frejelón is an species, found nowhere else on earth. endangered. From quizlet

Answer: plant cover provided food and protection for organisms riparian zone-buffers water, The Frejelón is an species, found nowhere else on earth. endangered.

Explanation:

Chole uses the S.M.A.R.T. guidelines to set a goal. She plans to set aside 15 hours to study each day so that she will earn straight A's on her report card. Which step did she overlook? (5 points) Attainable Realistic Specific Timely

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Answer: Realistic

Explanation: would you really be able to study for 15 hrs per day?, would those hrs have breaks between them?, and if you might have activity or things you might need to get done would that cut down on your time to study? These question come in range of making you realize whether it falls in the category of being realistic.

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