Question

Use the Integrating Factor method to solve: (dy)/(dt)+y=2 with y(0)=1. The integrating factor is of the form e^(kt). What is k ? You should find the solution is of the for: y(t)=A+Beᶜᵗ

Answer 1

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Answer 2

The integrating factor
`\(e^(kt)\)` is given by
`\(e^(kt) = e^(-t)\)`, which implies k = -1. Therefore, the solution to the differential equation
`\((dy)/(dt) + y = 2\)` with y(0) = 1 is
`\(y(t) = 2 - e^(-t)\)`.

Step-by-step explanation

• Identifying Integrating Factor:

We start with the differential equation
`\((dy)/(dt) + y = 2\)` and identify the integrating factor, denoted as
`\(e^(kt)\)`. In this case, the integrating factor is
`\(e^(kt) = e^(-t)\)` , implying k = -1.

`\[ \text{Integrating Factor: } e^(kt) = e^(-t) \]`

• Multiplying and Rearranging:

Multiply both sides of the differential equation by the integrating factor
`\(e^(-t)\)`:

`\[ e^(-t) (dy)/(dt) + e^(-t)y = 2e^(-t) \]`

Recognizing that the left side is the derivative of
`\(e^(-t)y\)` , we can rewrite the equation:

`\[ (d)/(dt)(e^(-t)y) = 2e^(-t) \]`

• Integration:

Integrate both sides with respect to t:

`\[ \int (d)/(dt)(e^(-t)y) \,dt = \int 2e^(-t) \,dt \]`

This leads to:

`\[ e^(-t)y = -2e^(-t) + C \]`

where C is the constant of integration.

• Solving for y:

Solve for y by multiplying through by \(e^t\):

`\[ y = -2 + Ce^(t) \]`

• Applying Initial Condition:

Apply the initial condition y(0) = 1 to find the value of C:

`\[ 1 = -2 + C \]`

`\[ C = 3 \]`

• Final Solution:

Substitute C back into the solution:

`\[ y(t) = 2 - e^(-t) \]`

This detailed process ensures a comprehensive understanding of how the solution was derived.