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vanitymdl
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Problem: Find the solution to the initial Value problem (differential equations)?
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^xHow does this look?
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^xHow does this look?