# Exact Equations

Differential equation M(x,y)dx+N(x,y)dy=0 is exact if there exist function f such that df=M(x,y)dx+N(x,y)dy .

In this case equation can be rewritten as df=0 which gives solution f=C.

Test for exactness: if M(x,y) and N(x,y) are continuous functions and have continuous first partial derivatives on some rectangle of the xy-plane, then differential equation is exact if and only if (partial M(x,y))/(partial y)=(partial N(x,y))/(partial x).

To solve this equation we use the facts that (partial f(x,y))/(partial x)=M(x,y) and (partial f(x,y))/(partial y)=N(x,y). We integrate first equation with respect to x to obtain f(x,y) through x and unknown function g(y). We then differentiate result with respect to y and use second equation. After this we find g(y) and thus f(x,y). Solution, as already stated above, is given as f(x,y)=C .

Example 1. Solve (x+sin(y))dx+(x cos(y)-2y)dy=0

Here M(x,y)=x+sin(y) , N(x,y)=xcos(y)-2y . Since (partial M)/(partial y)=cos(y) and (partial N)/(partial x)=cos(y) then (partial M)/(partial y)=(partial N)/(partial x) and differential equation is exact.

Thus, there exist function f such that (partial f)/(partial x)=M(x,y)=x+sin(y) and (partial f)/(partial y)=N(x,y)=xcos(y)-2y .

Integrate first equation with respect to x to obtain that f=int (x+sin(y))dx=1/2 x^2+xsin(y)+g(y) .

Note that here constant of integration can depend on y (because we integrate with respect to x).

Now diffetiate resulting equation with respect to y:

(partial f)/(partial y)=0+xcos(y)+g'(y) .

From another side (partial f)/(partial y)=N(x,y)=xcos(y)-2y

So, xcos(y)+g'(y)=xcos(y)-2y , or g'(y)=-2y . Integrating it we obtain that g(y)=-y^2+c_1 .

Thus,

f=1/2 x^2+xsin(y)-y^2+c_1 . Solution is given as 1/2 x^2 +xsin(y)-y^2+c_1=C

Or 1/2 x^2 +xsin(y)-y^2=c_2 where (c_2=C-c_1 ).

We obtain implicit solution and there is no way to find explicit solution.

Example 2. Solve y'=(2+ye^(xy))/(2y-xe^(xy)) .

First rewrite it into differential form: (dy)/(dx)=(2+ye^(xy))/(2y-xe^(xy)) or (2+ye^(xy))dx+(xe^(xy)-2y)dy=0 .

Here M(x,y)=2+ye^(xy) and N(x,y)=xe^(xy)-2y .

Since (partial M)/(partial y)=e^(xy)+xye^(xy) and (partial N)/(partial x)=e^(xy)+xye^(xy) then (partial M)/(partial y)=(partial N)/(partial x) and differential equation is exact.

So, exist function f such that (partial f)/(partial x)=M(x,y)=2+ye^(xy) and (partial f)/(partial y)=N(x,y)=xe^(xy)-2y . In previous example we took first equation and integrated it with respect to x. Actuall y we could take second equation and integrate it with respect to y. Final answer is the same. In this example we will take second equation and integrate it with respect to y:

f=int (xe^(xy)-2y)dy=e^(xy)-y^2+g(x) . Note that constant of integration depends on x, since we integrate with respect to y.

Now, differentiate resulting equation with respect to x:

(partial f)/(partial x)=y e^(xy)+g'(x)

From another side (partial f)/(partial x)=2+ye^(xy) , so ye^(xy)+g'(x)=2+ye^(xy) or g'(x)=2 .

Integrating with respect to x gives g(x)=2x+c_1 .

So, f=e^(xy)-y^2+g(x)=e^(xy)-y^2+2x+c_1 .

And solution is e^(xy)-y^2+2x+c_1=C or e^(xy)-y^2+2x=c_2 where (c_2=C-c_1) .

Sometimes, equation can be not exact, but it can be transformed into exact by multiplying equation by integrating factor.

So, I(x,y) is integrating factor for differential equation is exact if I(x,y)(M(x,y)dx+N(x,y)dy)=0 is exact.

There are 2 conditions that allow to find integrating factor easily:

If 1/N((partial M)/(partial y)-(partial N)/(partial x))=g(x) , a function of x alone, then I(x,y)=e^(int g(x)dx)

If 1/M((partial M)/(partial y)-(partial N)/(partial x))=h(y) , a function of x alone, then I(x,y)=e^(-int h(y)dy)

Example 3. Solve (y+1)dx-xdy=0

Here M(x,y)=y+1 and N(x,y)=-x . Since (partial M)/(partial y)=1 and (partial N)/(partial x)=-1 then (partial M)/(partial y)!=(partial N)/(partial x) and equation is not exact. However, note that 1/N((partial M)/(partial y)-(partial N)/(partial x))=1/(-x)(1-(-1))=-2/x=g(x) .

So, I(x,y)=e^(int -2/xdx)=e^(-2ln(x))=1/x^2 .

Multiplying differential equation by integrating factor gives: (y+1)/x^2dx-1/xdy=0.

This equation is exact, so exist function f such that df=(y+1)/x^2 dx-1/x dy . Using the fact that (partial f)/(partial x)=(y+1)/x^2, we find that f=int (y+1)/x^2 dx=-(y+1)/x+g(y) . Now, differentiating with respect to y gives: (partial f)/(partial y)=-1/x+g'(y) . From another side (partial f)/(partial y)=-1/x , so -1/x+g'(y)=-1/x or g'(y)=0 .

Solving it we obtain that g(y)=C_1 .

So, f(x,y)=-(y+1)/x+g(y)=-(y+1)/x+C_1 .

Therefore, solution is -(y+1)/x+C_1=C or y=c_2 x-1 where c_2=C_1-C .

Example 4. Solve 2xydx+y^2dy=0

Here M(x,y)=2xy and N(x,y)=y^2 . Since (partial M)/(partial y)=2x and (partial N)/(partial x)=0 then (partial M)/(partial y)!=(partial N)/(partial x) and differential equation is not exact.

However, note that 1/M((partial M)/(partial y)=(partial N)/(partial x))=1/(2xy)(2x-0)=1/y=h(y) .

So, integrating factor is I(x,y)=e^(-int 1/y dy)=e^(-ln(y))=1/y .

Multiplying differential equation by integrating factor yields 1/y(2xydx+y^2dy)=0 or 2xdx+ydy=0

This equation is exact. Using the fact that (partial f)/(partial x)=2x we have that f=int (2x)dx=x^2+h(y) . Differentiating last eqaution with respect to y gives: (partial f)/(partial y)=h'(y) .

From another side (partial f)/(partial y)=y , so h'(y)=y or h(y)=int ydy=1/2 y^2+c_1

So, f=x^2+h(y)=x^2+1/2 y^2+c_1 and solution is  x^2+1/2 y^2+c_1=C or x^2+1/2 y^2=c_2 where (c_2=C-c_1 ) .

Another case when there can be easily found integrating factor is case when M=yf(xy) and N=xg(xy) . In this case I(x,y)=1/(xM-yN) .

Example 5. Solve y(1-xy)dx+xdy=0 .

Equation is not exact, however, note that M(x,y) is in the form y(1-xy) and N(x,y)=x*1 ,so integrating factor is I(x,y)=1/(xy(1-xy)-xy)=-1/(xy)^2 . Multiplying by I(x,y) yields:

(xy-1)/(x^2y)dx-1/(xy^2)dy=0 . This equation is exact.

Using the fact that (partial f)/(partial x)=(xy-1)/(x^2y) we have that f=int ((xy-1)/(x^2 y))dx=ln(|x|)+1/(xy)+g(y) .

Differentiating with respect to y gives: (partial f)/(partial y)=-1/(xy^2)+g'(y) . From another side (partial f)/(partial y)=-1/(xy^2) .

So, g'(y)=0 or g(y)=c_1 .

So, f(x,y)=ln(|x|)+1/(xy)+c_1 .

Finally, solution is ln(|x|)+1/(xy)+c_1=C or ln(|x|)+1/(xy)=c_2 where c_2=C-c_1 .

Below is the list of common integrating factors:

 Group of terms I(x,y) Exact differential dg(x,y) ydx-xdy -1/x^2 -(ydx-xdy)/x^2=d(y/x) ydx-xdy 1/y^2 (ydx-xdy)/y^2=d(x/y) ydx-xdy -1/(xy) -(ydx-xdy)/(xy)=d(ln(y/x)) ydx-xdy -1/(x^2+y^2) -(ydx-xdy)/(x^2+y^2)=d(arctan(y/x)) ydy+xdx 1/(xy) (ydy+xdx)/(xy)=d(ln(xy)) ydy+xdx 1/(xy)^n,n>1 (ydy+xdx)/(xy)^n=-d(1/((n-1)(xy)^(n-1))) ydy+xdx 1/(x^2+y^2) (ydy+xdx)/(x^2+y^2)=d(1/2 ln(x^2+y^2)) ydy+xdx 1/(x^2+y^2)^n, n>1 (ydy+xdx)/(x^2+y^2)^n=-d(1/(2(n-1)(x^2+y^2)^(n-1))) aydx+bxdy (a and b constants) x^(a-1)y^(b-1) x^(a-1)y^(b-1)(aydx+bxdy)=d(x^ay^b)

In general, integrating factors are difficult to uncover. If a differential equation doesn't have a one of the forms given above, then a search for an integrating factor likely will not be successful, and other methods of solution should be used.