How to Calculate Integrating Factor

Think about this problem: dy/dx + 4y = 16 where y(0)=9

Assume that we already know the integrating factor u(x) and we just multiple the equation by it so we get:

u(x)* dy/dx + 4u(x)y = 16u(x)   (Equation 1)

We know that the derivative of u(x)y witht respect to x is d(u(x)y)/dx = u(x)*dy/dx +y du(x)/dx  (Equation 2)

Combing Equation 1 and 2 will give you this:

du(x)/dx =4u(x)

And that is same as this:

(du(x)/dx) /u(x)=4

Multiply dx on both sides so we get:

du(x) /u(x)=4 dx

Perform integration and we can attain this:

ln |u(x)|=4x+b  (b is called constant of integration here)

Raise e to the power of both sides of the equation and then we get:

u(x)= c e^(4x)     (The formation of the constant c is caused by the existance of the constant b, where c=e^b)

Therefore the integrating factor is just c e^(4x)

Remember the first step u(x)* dy/dx + 4u(x)y = 16u(x) ? We are now replacing the u(x) with c e^(4x):

c e^(4x)* dy/dx + 4* c e^(4x)* y = 16* c e^(4x)

The c cancels out:

e^(4x)* dy/dx + 4 e^(4x)* y = 16e^(4x)

We now can see that the left side of the equation is actually equal to the derivative of e^(4x) y, so

d (e^(4x) y)/dx=16e^(4x)

Multiply dx on both sides:

 d (e^(4x) y)=16e^(4x) dx

Now do integraion:

e^(4x) y=4 e^(4x) + C

We are almost there, now y=4+C*e^(-4x))

Because y(0)=9 is given so that means C=5 according to the equation we got above.

Therefore the answer is y=4+5*e^(-4x))



  • Subject : Math
  • Topic : Calculus
  • Posted By : Jason

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