How Is the Integrating Factor Used to Solve Initial Value Differential Problems?

[vanitymdl]

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?

 

[Curious3141]

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^x


How does this look?

The first thing you should do is to substitute your solution back into the d.e. Does it solve the equation?

Is this $y' = -y + e^{-2x}$ or is this $y' - y = e^{-2x}$ your equation?

Because in the problem statement, you gave the former, but you solved the latter. The solution for the latter is correct, but if you made a simple algebraic error in bringing the $y$ over, then the solution is obviously wrong.

 

[LCKurtz]

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

Isn't C included in the "everything"?

 

[Curious3141]

Isn't C included in the "everything"?

Yes, it should be. The reason it didn't affect the answer is because $e^0 = 1$.

 

[vanhees71]

Why not using the standard way of solving linear equations, i.e., first solving the homogeneous equation
y'=-y
and then using the ansatz of the variation of the constant?

 

[Curious3141]

Why not using the standard way of solving linear equations, i.e., first solving the homogeneous equation
y'=-y
and then using the ansatz of the variation of the constant?

The integrating factor is the usual elementary method that's taught for questions like this, I think.

I was about to suggest using a Laplace transform, which gives a quick algebraic solution in an initial value problem like this, but thought better of it, simply because it's unlikely the thread starter has covered it yet.