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y'=-y+e^-2x

y(0) = 3

Attempt: y' -y = e^(-2x) ----- (1)

Linear equation of first order of form y' + y p(x) = q(x)

p(x)= -1

q(x) = e^(-2x)

Find the integrating factor e^∫p(x) dx = e^∫ - dx = e^-x

Multiply equation (1) by the integrating factor e^-x

y' e^-x - y e^-x = e^(-3x) ----- (2)

The left-hand side of (2) is the derivative of y times the integrating factor or (y e^-x)'

(y e^-x)' = e^(-3x) ----- (3)

Integrate both sides of (3)

y e^-x = ∫ e^(-3x)

y e^-x = (-1/3) e^(-3x) + C

multiply everything by e^x

y = (-1/3) e^(-2x) + C

y(0)=3

3 = (-1/3) e^0 + C

C = 3+1/3 = 10/3

y = (-1/3) e^(-2x) + (10/3) e^x

How does this look?