ABSTRACT
M ( x , y ) d x + N ( x , y ) d y = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203743102/9cca8f32-4688-4a1e-80c2-a6c9068f5153/content/eq445.tif"/>
If the equation can be put in the form A(x)dx + B(y)dy = 0, it is separable and the solution follows by integration: ∫ A ( x ) d x + ∫ B ( y ) d y = C ; https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203743102/9cca8f32-4688-4a1e-80c2-a6c9068f5153/content/eq446.tif"/> thus, x(1 + y 2)dx + ydy = 0 is separable since it is equivalent to xdx + ydy/(1 + y 2)=0, and integration yields x 2 / 2 + 1 2 log ( 1 + y 2 ) + C = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203743102/9cca8f32-4688-4a1e-80c2-a6c9068f5153/content/eq447.tif"/> .
If M(x, y) and N(x, y) are homogeneous and of the same degree in x and y, then substitution of υx for y (thus, dy = υ dx + x dυ) will yield a separable equation in the variables x and y. [A function such as M(x, y) is homogeneous of degree n in x and y if M(cx, cy) = cnM(x, y).]For example, (y−2x)dx + (2y + x)dy has M and N each homogeneous and of degree one so that substitution of y = υx yields the separable equation 2 x d x + 2 υ + 1 υ 2 + υ − 1 d υ = 0. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203743102/9cca8f32-4688-4a1e-80c2-a6c9068f5153/content/eq448.tif"/>
If M(x, y)dx + N(x, y)dy is the differential of some function F(x, y),then the given equation is said to be exact. A necessary and sufficient 106condition for exactness is ∂ M / ∂ y = ∂ N / ∂ x https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203743102/9cca8f32-4688-4a1e-80c2-a6c9068f5153/content/eq449.tif"/> . When the equation is exact, Fis found from the relations ∂F/∂x = M and ∂F/∂y = N, and the solution is F(x, y = C (constant). For example, (x 2 + y)dy + (2xy − 3x 2)dx is exact since ∂M/∂y = 2x and ∂N/∂x = 2x. F is found from ∂F/∂x = 2xy − 3x 3 and ∂F/∂y = x 2 + y.From the first of these, F = x 2 y−x 3 + ϕ(y); from the second, F = x 2 y+y 2/2 + ¥(x). It follows that F = x 2 y = x 3 + y 2/2, and F = C is the solution.
Linear, order one in y: Such an equation has the form dy + P(x)ydx = Q(x)dx. Multiplication by exp[∫P(x)dx] yields d [ y exp ( ∫ P d x ) ] = Q ( x ) exp ( ∫ P d x ) d x . https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780203743102/9cca8f32-4688-4a1e-80c2-a6c9068f5153/content/eq450.tif"/>
For example, dy + (2/x)ydy = x 2 dx is linear in y. P(x) = 2/x, so ∫P dx = 2 ln x = ln x 2, and exp(∫P dx) = x 2. Multiplication by x 2 yields d(x 2 y = x 4 dx, and integration gives the solution x 2 y = x 5/5 + C.
