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Chris L T521 · #2 May 13th, 2008, 12:30 PM

3. The Integrating Factor

The method of the integrating factor is used when we have differential equations in the form $\frac{dy}{dx}+P(x)y=Q(x)$ . Multiplying the equation through by the integrating factor $e^{\int P(x)\,dx}$ , we would have the equation ${\left[e^{\int P(x)\,dx}y\right]}^{/}=Q(x)e^{\int P(x)\,dx}$ . Integrating both sides and solving for y, we get:

$y=e^{-\int P(x)\,dx}{\left[\int Q(x)e^{\int P(x)\,dx}\,dx\right]}$ .

Let us go through an easy example, and then a challenging one.

Example 5:

Solve $x\frac{dy}{dx}+y=3xy$ ; $y(1)=0$ .

$x\frac{dy}{dx}+y=3xy \implies x\frac{dy}{dx}+(1-3x)y=0$

In order to apply the integrating factor, the coefficient of $\frac{dy}{dx}$ must be equal to 1.

$x\frac{dy}{dx}+(1-3x)y=0 \implies \frac{dy}{dx}+{\left(\frac{1}{x}-3\right)}y=0$ .

Now find the integrating factor:

$\rho(x)=e^{\int P(x)\,dx}=e^{\int {\left(\frac{1}{x}-3\right)}\,dx}=e^{lnx-3x}=xe^{-3x}$ .

Multiplying through, we should get:

${\left[xe^{-3x}y\right]}^{/}=0$

Integrating, we find that:

$xe^{-3x}y=C$

Imposing the initial condition $y\left(1\right)=0$ , we see that $e^{-3}\cdot0=C\implies C=0$

Therefore, the solution to the differential equation is $y=0\cdot x^{-1}e^{3x}\implies \color{red}\boxed{y \equiv 0}$ .

Example 6:

Solve $(x^2+1)\frac{dy}{dx}+3x^3y=6xe^{-\frac{3}{2}x^2}$ ; $y(0)=1$ .

Divide through by $x^2+1$ :

$\frac{dy}{dx}+\frac{3x^3}{x^2+1}y=\frac{6xe^{-\frac{3}{2}x^2}}{x^2+1}$ .

Now find the integrating factor:

$\rho(x)=e^{\int P(x)\,dx}=e^{\int \frac{3x^3}{x^2+1}\,dx}$

Apply long division to simplify the integrand:

(Verify): $\rho(x)=e^{\int{\left(3x-\frac{3x}{x^2+1}\right)}\,dx}=e^{\frac{3}{2}x^2-\frac{3}{2}ln(x^2+1)}=(x^2+1)^{-\frac{3}{2}}e^{\frac{3}{2}x^2}$ .

Multiplying through by the integrating factor, we should get:

(Verify): ${\left[(x^2+1)^{-\frac{3}{2}}e^{\frac{3}{2}x^2}y\right]}^{/}=\frac{6x}{(x^2+1)^{\frac{5}{2}}}$

Integrating both sides and then solving for y, we get:

$(x^2+1)^{-\frac{3}{2}}e^{\frac{3}{2}x^2}y=-\frac{2}{(x^2+1)^{\frac{5}{2}}}+C \implies y=e^{-\frac{3}{2}x^2}{\left(C(x^2+1)^{\frac{3}{2}}-2\right)}$ .

Now apply the initial condition: y(0)=1:

$1=C-2 \implies C=3$ .

Therefore, our particular solution will be:

$\color{red}\boxed{y=e^{-\frac{3}{2}x^2}{\left(3(x^2+1)^{\frac{3}{2}}-2\right)}}$ .

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4. Exact Equations

In order to use the technique to solve exact equations, the equations must be in the form:

$M(x,y)dx+N(x,y)dy=0$ ,

And they must satisfy this one condition:

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

If this relationship is true, we'll continue on with this technique. If its not true, we will resort to 2 other possible techniques which will be discussed later.

When we go about solving this, we should make known that $\frac{\partial f}{\partial x}=M(x,y)$ and that $\frac{\partial f}{\partial y}=N(x,y)$ .

Step one: find f(x,y). You can do it two ways, but I will do it this way because its the most common way.

$\frac{\partial f}{\partial x}=M(x,y) \implies \int\frac{\partial f}{\partial x}\,dx=\int M(x,y)\,dx \implies f(x,y)=\int M(x,y)\,dx + g(y)$

Step 2: Find g(y). To do this, partially differentiate f(x,y) with respect to y.

$\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}\int M(x,y)\,dx+g^{/}(y)$

Since $N(x,y)=\frac{\partial f}{\partial y}$ ,

$N(x,y)=\frac{\partial}{\partial y}\int M(x,y)\,dx+g^{/}(y)$

Solving for $g^{/}(y)$ , we get:

$N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\,dx=g^{/}(y)$ .

Integrate to find g(y):

$g(y)=\int{\left(N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\,dx\right)}\,dy$ .

Step 3: write solution in general form.

The general solution of an exact equation will have the form:

$f(x,y)=C$ .

Since $f(x,y)=\int M(x,y)\,dx + \int{\left(N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\,dx\right)}\,dy$ , the general solution will be:

$\int M(x,y)\,dx + \int{\left(N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\,dx\right)}\,dy=C$ .

Example 7:

Solve $(cosx+lny)dx+{\left(\frac{x}{y}+e^y\right)}dy=0$ .

$M(x,y)=cosx+lny$
$N(x,y)=\frac{x}{y}+e^y$

Test for exactness:

$\frac{\partial M}{\partial y}=\color{red}\frac{1}{y}$
$\frac{\partial N}{\partial x}=\color{red}\frac{1}{y}$

They are equal, so they are exact equations.

Find f(x,y):

$f(x,y)=\int (cosx+lny)\,dx \implies f(x,y)=sinx+xlny+g(y)$ .

Now find g(y):

$\frac{\partial f}{\partial y}=\frac{x}{y}+g^{/}(y)$ .

Since $N(x,y)=\frac{\partial f}{\partial y}$ ,

$\frac{x}{y}+e^y=\frac{x}{y}+g^{/}(y) \implies g^{/}(y)=e^y \implies g(y)=e^y$

Therefore,

$f(x,y)=sinx+xlny+e^y$ .

Therefore, the general solution is:

$\color{red}\boxed{sinx+xlny+e^y=C}$ .

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