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Now, let's consider the following example,

Â which is first order,

Â linear, non-homogeneous differential equation.

Â Say, x times y prime + (1 + x)y

Â is equal to e to_the_negative x times cosine 3x.

Â First, note that the differential equation is not always standard form.

Â In other words, the leading coefficients of differential equation,

Â that is the coefficient of y prime,

Â is not 1 but x.

Â So, divide it through by x. Divide it through the equation by x,

Â then y prime plus (1 + x) over x and y,

Â that is equal to e to_the_negative x,

Â over x and times cosine 3 of x.

Â Now, we can recognize from this expression what is p?

Â Be careful p is not 1+ x ,

Â p must be, (1+ x) over x.

Â What is q?

Â q(x) is, e to_the_negative x,

Â over x, times the cosine 3 of x.

Â What I'm claiming is,

Â this part is the the little p(x).

Â And that part is equal to little q(x).

Â So, what is the integrating factor?

Â Integrating factor mu of x is equal to,

Â e to the integral px, dx.

Â What is p? (1 + x) over x.

Â So, there we have e to the integral (1 + x) over x, dx.

Â So, what is the anti-derivative of (1+ x) over x?

Â That is the same as the 1 over x plus 1.

Â So you have- actually,

Â this is equal to e to the log of x and plus x.

Â By taking the integral constant to be equal to zero,

Â so what does that mean?

Â This is equal to, e to_the_x times, e to the log of x,

Â that is equal to x times e to_the_x.

Â This is our integrating factor.

Â This is our integrating factor.

Â Multiply this integrating factor,

Â mu of x, on both sides of this equation.

Â So, what we needed to do is for this whole equation,

Â multiply it by, x times e to_the_x.

Â Multiply.

Â Then what's this left inside?

Â It should to be, x times e to_the_x times

Â y and derivative of e. That should be equal to,

Â x times e to_the_x times e to_the_negative x,

Â over x times cosine 3x.

Â This over x canceled out,

Â e to_the_x, e to_the_negative x canceled out.

Â So finally you get,

Â this times that, is equal to cosine 3 of x.

Â That's the equation we have down here.

Â Derivative of x times e to_the_x of y, is equal to cosine 3 of x. What does that mean?

Â x times e to_the_x y is equal to y.

Â So, from this equation down there,

Â you have x times e to_the_x,

Â times y is equal to,

Â anti-derivative of cosine 3x, dx.

Â This is equal to one-third.

Â And the sine of 3 of x plus,

Â arbitrary constant c. Then finally divided through by,

Â x times e to_the_x,

Â you are going to get to our final solution,

Â y is equal to 1 over x, times e to_the_x,

Â times one-third sine 3x + c. Now,

Â let's introduce one another class.

Â So called the exact differential equation.

Â First, for any function,

Â F(x, y) with continuous first partial derivatives,

Â we call dF.

Â We call dF. Which is the same as,

Â dF over dx times dx,

Â plus dF over dy times dy.

Â We call this expression,

Â dF, the total differential of capital F. Then,

Â along any level curve,

Â F( x, y ) = constant,

Â we should have, the total differential,

Â dF, must be equal to zero.

Â Because the total differential of any constant is equal to zero.

Â So, we have such an equation down here.

Â dF over dx times dx + dF over dy times dy is equal to 0.

Â Or equivalently, if you wanted to express it,

Â as y prime is equal to something,

Â then, you can solve it,

Â dy over dx is equal to;

Â negative dF over dx over dF over dy.

Â Which is the first order differential equation.

Â Conversely, for a differential equation,

Â given here, M(x, y) dx + N( x,

Â y )dy = 0.

Â This is a first order differential equation.

Â If this left hand side,

Â is a total differential of F(x, y ),

Â then the differential equation can be written as,

Â because this left hand side is a total differential of capital F,

Â so you can write is as,

Â dF(x, y) = 0.

Â We can solve with this a simple differential equation very easily.

Â It means simply, F( x,

Â y) is equal to arbitrary constant c. Arbitrary constant c down there.

Â So, this is equal to F(x,

Â y) = c, is a general solution to this first order differential equation.

Â If this left hand side is known to be the total differential of capital F(x, y).

Â That motivates us to the following definition.

Â A force to the differential equation,

Â M(x,y) dx + N(x,y) dy = 0,

Â we call it to be exact,

Â in a plane region R if there is a function capital F(x,

Â y) such that dF over dx is equal to M(x,

Â y) and dF over dy is equal to N(x, y).

Â For all (x, y) in the region R. In other words what?

Â This left hand side of the differential equation,

Â is the total differential of capital F(x,

Â y) in the region R. Then we call,

Â the given differential equation to be exact.

Â If it is exactly in the whole plane,

Â R squared, if R is- the region R,

Â is the whole plane R_two, R squared.

Â Then, we simply say that,

Â the given differential equation, M(x,

Â y) dx + N(x, y) dy = 0, is exact.

Â