Answer:
Option A. [tex]y = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}[/tex]
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:
[tex]y''=e^{-2t}+10e^{4t}[/tex]
[tex]\frac{dy'}{dt}=e^{-2t}+10e^{4t}[/tex]
[tex]dy'=[e^{-2t}+10e^{4t}]dt[/tex]
[tex]\int dy'=\int [e^{-2t}+10e^{4t}]dt[/tex]
Applying various properties of integration:
[tex]\int dy'=\int e^{-2t} dt + \int 10e^{4t}dt\\\int dy'=\int e^{-2t} dt + 10\int e^{4t}dt[/tex]
Prepare for integration by u-substitution
[tex]\int dy'=\int e^{u_1} dt + 10\int e^{u_2}dt[/tex], letting [tex]u_1=-2t[/tex] and [tex]u_2=4t[/tex]
Find dt in terms of [tex]u_1 \text{ and } u_2[/tex]
[tex]u_1=-2t\\du_1=-2dt\\-\frac{1}{2}du_1=dt[/tex] [tex]u_2=4t\\du_2=4dt\\\frac{1}{4}du_2=dt[/tex]
[tex]\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)[/tex]
Using the Exponential rule (don't forget your constant of integration):
[tex]y'=-\frac{1}{2} e^{u_1} + 10 *\frac{1}{4}e^{u_2} +C_1[/tex]
Back substituting for [tex]u_1 \text{ and } u_2[/tex]:
[tex]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\\[/tex]
Finding the constant of integration
Given initial condition [tex]y'(0)=0[/tex]
[tex]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\\[/tex]
The first derivative with the initial condition applied: [tex]y'(t)=-\frac{1}{2} e^{-2t} + \frac{5}{2}e^{4t} -2\\[/tex]
Round 2:
Integrate again:
[tex]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\\[/tex]
[tex]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[/tex]
Finding the constant of integration :
Given initial condition [tex]y(0)=1[/tex]
[tex]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[/tex]
So, [tex]y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}[/tex]
Checking the solution
[tex]y(t) = \frac{1}{4} e^{-2t} + \frac{5}{8} e^{4t} - 2 t + \frac{1}{8}[/tex]
This matches our initial conditions here [tex]y(0) = \frac{1}{4} e^{-2(0)} + \frac{5}{8} e^{4(0)} - 2 (0) + \frac{1}{8} = 1[/tex]
Going back to the function, differentiate:
[tex]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}]'[/tex]
Apply Exponential rule and chain rule, then power rule
[tex]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[/tex]
This matches our first order step and the initial conditions there.
[tex]y'(0) = -\frac{1}{2} e^{-2(0)} + \frac{5}{2} e^{4(0)} - 2=0[/tex]
Going back to the function y', differentiate:
[tex]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]'[/tex]
Applying the Exponential rule and chain rule, then power rule
[tex]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}[/tex]
So our proposed solution is a solution to the differential equation, and satisfies the initial conditions given.
