Solve the following initial value problem explicitly:
y'= ye^-x , y(0)=1
How do you solve this problem? I'm really confused!
y' = y e^(-x)
Separate variables to get
dy/y = e^(-x)dx
Integrate
ln(y) = -e^(-x) + C'
ln(y) = -e^(-x) + C'
raise to power to base e
e^(ln(y))=e^(-e(-x)+C')
y = Ce^(e^(-x))
(the negative sign and C' are absorbed in the new constant C)
y(0)=1
=>
1 = Ce^(-e^0) = Ce
C=e
=>
y=e^(-e^(-x) +1)
(Note: it's easier if you always post with the same name, Leah, Grace, and Tania is too confusing.)
To solve this initial value problem, we will use the method of separation of variables.
Step 1: Write the given differential equation in the form dy/dx = f(x)g(y). In this case, we have y' = y * e^(-x).
Step 2: Separate the variables by moving all terms involving y to one side and all terms involving x to the other side. Divide both sides by y * e^(-x) to isolate the y terms on one side and the x terms on the other side.
dy / y = e^(-x) dx
Step 3: Integrate both sides of the equation with respect to their respective variables.
∫(1/y) dy = ∫e^(-x) dx
Step 4: Evaluate the integrals using the appropriate integration rules:
ln|y| = -e^(-x) + C
Here, C is the constant of integration.
Step 5: Solve for y by exponentiating both sides of the equation:
|y| = e^(-e^(-x) + C)
Step 6: Remove the absolute value by considering the positive and negative cases:
Case 1: y = e^(-e^(-x) + C)
Case 2: y = -e^(-e^(-x) + C)
Step 7: Apply the initial condition, y(0) = 1, to find the value of the constant C.
For Case 1: 1 = e^(-e^0 + C) = e^(C)
Thus, C = ln(1) = 0.
So the solution for Case 1 is y = e^(-e^(-x)).
For Case 2: 1 = -e^(-e^0 + C) = -e^(C)
There is no solution for Case 2 that satisfies the initial condition.
Therefore, the explicit solution to the given initial value problem is y = e^(-e^(-x)).