Advanced Placement (AP)
21.10.2022 13:42
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Solve the following differential equation with initial conditions: y =e^-2t+10e^4t

Solve the following differential equation with initial conditions: y''=e^-2t+10e^4t ; y(0)=1, y'(0)=0
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devnnn44
devnnn44
5,0(30 marks)

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.

jdanstudy9528
jdanstudy9528
4,6(71 marks)
maeve Kinkead? she is an American soap opera actress

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