Question

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

Answer 1

Option A.
`y = (1)/(4) e^(-2t) + (5)/(8) e^(4t) - 2 t + (1)/(8)`

Step-by-step 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)`

`(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\\-(1)/(2)du_1=dt`
`u_2=4t\\du_2=4dt\\(1)/(4)du_2=dt`

`\int dy'=\int e^(u_1) dt + 10\int e^(u_2)dt\\\int dy'=\int e^(u_1) (-(1)/(2) du_1) + 10\int e^(u_2) ((1)/(4) du_2)\\\int dy'=-(1)/(2) \int e^(u_1) (du_1) + 10 *(1)/(4) \int e^(u_2) (du_2)`

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

`y'=-(1)/(2) e^(u_1) + 10 *(1)/(4)e^(u_2) +C_1`

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

`y'=-(1)/(2) e^((-2t)) + 10 *(1)/(4)e^((4t)) +C_1\\y'=-(1)/(2) e^(-2t) + (5)/(2)e^(4t) +C_1\\`

Finding the constant of integration

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

`y'(t)=-(1)/(2) e^(-2t) + (5)/(2)e^(4t) +C_1\\0=y'(0)=-(1)/(2) e^(-2(0)) + (5)/(2)e^(4(0)) +C_1\\0=-(1)/(2) (1) + (5)/(2)(1) +C_1\\-2=C_1\\`

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

Round 2:

Integrate again:

`y' =-(1)/(2) e^(-2t) + (5)/(2)e^(4t) -2\\(dy)/(dt) =-(1)/(2) e^(-2t) + (5)/(2)e^(4t) -2\\dy =[-(1)/(2) e^(-2t) + (5)/(2)e^(4t) -2]dt\\\int dy =\int [-(1)/(2) e^(-2t) + (5)/(2)e^(4t) -2]dt\\\int dy =\int -(1)/(2) e^(-2t) dt + \int (5)/(2)e^(4t) dt - \int 2 dt\\\int dy = -(1)/(2) \int e^(-2t) dt + (5)/(2) \int e^(4t) dt - 2 \int dt\\`

`y = -(1)/(2) * -(1)/(2) e^(-2t) + (5)/(2) * (1)/(4) e^(4t) - 2 t + C_2\\y(t) = (1)/(4) e^(-2t) + (5)/(8) e^(4t) - 2 t + C_2`

Finding the constant of integration :

Given initial condition
`y(0)=1`

`1=y(0) = (1)/(4) e^(-2(0)) + (5)/(8) e^(4(0)) - 2 (0) + C_2\\1 = (1)/(4) (1) + (5)/(8) (1) - (0) + C_2\\1 = (7)/(8) + C_2\\(1)/(8)=C_2`

So,
`y(t) = (1)/(4) e^(-2t) + (5)/(8) e^(4t) - 2 t + (1)/(8)`

Checking the solution

`y(t) = (1)/(4) e^(-2t) + (5)/(8) e^(4t) - 2 t + (1)/(8)`

This matches our initial conditions here
`y(0) = (1)/(4) e^(-2(0)) + (5)/(8) e^(4(0)) - 2 (0) + (1)/(8) = 1`

Going back to the function, differentiate:

`y' = [(1)/(4) e^(-2t) + (5)/(8) e^(4t) - 2 t + (1)/(8)]'\\y' = [(1)/(4) e^(-2t)]' + [(5)/(8) e^(4t)]' - [2 t]' + [(1)/(8)]'\\y' = (1)/(4) [e^(-2t)]' + (5)/(8) [e^(4t)]' - 2 [t]' + [(1)/(8)]'`

Apply Exponential rule and chain rule, then power rule

`y' = (1)/(4) e^(-2t)[-2t]' + (5)/(8) e^(4t)[4t]' - 2 [t]' + [(1)/(8)]'\\y' = (1)/(4) e^(-2t)(-2) + (5)/(8) e^(4t)(4) - 2 (1) + (0)\\y' = -(1)/(2) e^(-2t) + (5)/(2) e^(4t) - 2`

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

`y'(0) = -(1)/(2) e^(-2(0)) + (5)/(2) e^(4(0)) - 2=0`

Going back to the function y', differentiate:

`y' = -(1)/(2) e^(-2t) + (5)/(2) e^(4t) - 2\\y'' = [-(1)/(2) e^(-2t) + (5)/(2) e^(4t) - 2]'\\y'' = [-(1)/(2) e^(-2t)]' + [(5)/(2) e^(4t)]' - [2]'\\y'' = -(1)/(2) [e^(-2t)]' + (5)/(2) [e^(4t)]' - [2]'`

Applying the Exponential rule and chain rule, then power rule

`y'' = -(1)/(2) e^(-2t)[-2t]' + (5)/(2) e^(4t)[4t]' - [2]'\\y'' = -(1)/(2) e^(-2t)(-2) + (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

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

Step-by-step 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\\\\&=-(1)/(2)e^(-2t)+10\cdot (1)/(4)e^(4t)+C\\\\&=-(1)/(2)e^(-2t)+(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\\\\-(1)/(2)e^(-2(0))+(5)/(2)e^(4(0))+C&=0\\\\-(1)/(2)(1)+(5)/(2)(1)+C&=0\\\\-(1)/(2)+(5)/(2)+C&=0\\\\2+C&=0\\\\C&=-2\end{aligned}`

Therefore, the equation for y' is:

`y'=-(1)/(2)e^(-2t)+(5)/(2)e^(4t)-2`

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

`\begin{aligned}\displaystyle y=\int y'&=\int -(1)/(2)e^(-2t)+(5)/(2)e^(4t)-2\;dt\\\\&=-(1)/(2)\int e^(-2t)\;dt+(5)/(2)\int e^(4t)\;dt-\int 2\;dt\\\\&=-(1)/(2) \cdot -(1)/(2)e^(-2t)+(5)/(2)\cdot (1)/(4)e^(4t)-2t+C\\\\&=(1)/(4)e^(-2t)+(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\\\\(1)/(4)e^(-2(0))+(5)/(8)e^(4(0))-2(0)+C&=1\\\\(1)/(4)(1)+(5)/(8)(1)-0+C&=1\\\\(1)/(4)+(5)/(8)+C&=1\\\\(7)/(8)+C&=1\\\\C&=(1)/(8)\end{aligned}`

Therefore the equation for y is:

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