Integrating factor technique

Integrating factor technique


Let’s suppose we wish to solve a differential problem of the form:

M(x,y)+N(x,y)dydx=0M(x,y)+N(x,y) \frac{dy}{dx}=0

But we cannot do the separable equations, and also MyNxM_y \neq N_x, so we cannot use exact equations.

But what if we could multiply the whole differential equation by some new equation that would make this problem exact? Let’s suppose there exists some sort of function that can do this trick. This function could be a function of xx or possibly some function of yy. Let’s suppose that the function that does this trick is μ(x)\mu(x).

μ(x)[M(x,y)+N(x,y)dydx=0]\mu(x)*[M(x,y)+N(x,y) \frac{dy}{dx}=0]

μ(x)M(x,y)+μ(x)N(x,y)dydx=0\Longrightarrow \mu(x)M(x,y)+\mu(x)N(x,y) \frac{dy}{dx}=0

And the whole goal of this is to have [μ(x)M(x,y)]y=[μ(x)N(x,y)]x[\mu(x)M(x,y) ]_y=[\mu(x)N(x,y) ]_x, so we can use our Exact Equations Method.

So if we can choose a μ(x)\mu(x) such that [μ(x)M(x,y)]y=[μ(x)N(x,y)]x[\mu(x)M(x,y) ]_y=[\mu(x)N(x,y) ]_x then choose this μ(x)\mu(x) and multiply the original equation by it:

μ(x)M(x,y)+μ(x)N(x,y)dydx=0\mu(x)M(x,y)+\mu(x)N(x,y) \frac{dy}{dx}=0

Now just solve this using the Exact Equation Method.

As we multiplied the entire equation by μ(x)\mu(x) every solution to

Where μ(x)0\mu(x)\neq 0 will also be a solution to

M(x,y)+N(x,y)dydx=0M(x,y)+N(x,y) \frac{dy}{dx}=0

Which was our original problem.

e.g. If we had an equation 2x=102x=10 and multiplied the whole equation by 2x+12x+1 (which could be our μ(x)\mu(x), then we will have



Note how x=5x=5 is a solution to both 2x=102x=10 and 4x2+2x=20x+104x^2+2x=20x+10. The only extra solution we picked up was x=12x=-\frac{1}{2}, which is the case where μ(x)=0\mu(x)=0.
  • Introduction
    What is the Integrating Factor Technique?

  • 1.
    Solving Differential Equation Using the Integrating Factor Technique
    Solve the following differential equation:

    4x+2y+2dydx=04x+2y+2 \frac{dy}{dx}=0

  • 2.
    Find a solution to the following first-order differential equation:

    y3ty=22ty'-\frac{3}{t} y=2- \frac{2}{t}

    With initial values y(1)=12y(1)=\frac{1}{2}