Questions: Lesson 1 Introduction: Problem 3 (1 point) It is easy to check that for any value of c, the function y = c e^(-2 x) + e^(-x) is solution of equation y' + 2 y = e^(-x) Find the value of c for which the solution satisfies the initial condition y(5) = 5. c =

Lesson 1 Introduction: Problem 3
(1 point)

It is easy to check that for any value of c, the function
y = c e^(-2 x) + e^(-x)
is solution of equation
y' + 2 y = e^(-x)

Find the value of c for which the solution satisfies the initial condition y(5) = 5.
c =

Transcript text: Lesson 1 Introduction: Problem 3 (1 point) It is easy to check that for any value of $c$, the function \[ y=c e^{-2 x}+e^{-x} \] is solution of equation \[ y^{\prime}+2 y=e^{-x} \] Find the value of $c$ for which the solution satisfies the initial condition $y(5)=5$. \[ c=\square \]

Solution

Solution Steps

Step 1: Define the Function

We start with the function given in the problem: \[ y = c e^{-2x} + e^{-x} \]

Step 2: Substitute the Initial Condition

We substitute $x = 5$ into the function to apply the initial condition $y(5) = 5$: \[ y(5) = c e^{-10} + e^{-5} \]

Step 3: Set Up the Equation

We set the expression equal to 5: \[ c e^{-10} + e^{-5} = 5 \]

Step 4: Solve for $c$

Rearranging the equation to isolate $c$: \[ c e^{-10} = 5 - e^{-5} \] Now, we can solve for $c$: \[ c = \frac{5 - e^{-5}}{e^{-10}} = (5 - e^{-5}) e^{10} \]

Step 5: Simplify the Expression

This simplifies to: \[ c = 5 e^{10} - e^{5} \]