Question

D.E. Help.

Teach me how can I solve these 3 problems, please. :<
1. y(4+y)dx - 2(x^2-y)dy = 0
2. 2(x-y+2)dx + x(x-2y+2)dy = 0
3. y(2x+y-2)dx - 2(x+y)dy = 0

Answer 1

@oldrin.bataku , sorry man. I still didn't get it. ._.

Answer 2

(1)$$y(4+y)dx-2(x^2-y)dy=0$$if you check you'll see this is not an exact equation:$$\frac{\partial}{\partial y}y(4+y)=4+2y,\frac{\partial}{\partial x}2(x^2-y)=4x-2y\\4+2y\ne4x-2y$$

Answer 3

now imagine we had a integration factor \(u\) that turned our equation into an exact one:$$uy(4+y)\,dx-2(x^2-y)u\,dy=0$$which gives the mixed partials:$$\frac{\partial}{\partial y}uy(4+y)=(4+2y)u+y(4+y)\frac{\partial u}{\partial y}\\\frac{\partial}{\partial x}2u(x^2-y)=(4x-2y)u+2(x^2-y)\frac{\partial u}{\partial x}$$since we want it to be exact we have:$$(4+2y)u+y(4+y)\frac{\partial u}{\partial y}=(4x-2y)u+2(x^2-y)\frac{\partial u}{\partial x}$$getting one \(u\) term alone on one side:$$(4x-2y-4-2y)u=y(4+y)\frac{\partial u}{\partial y}-2(x^2-y)\frac{\partial u}{\partial x}\\(4x-4y-4)u=u=y(4+y)\frac{\partial u}{\partial y}-2(x^2-y)\frac{\partial u}{\partial x}$$

Answer 4

the general idea is that we want to solve for \(u\) here but unfortunately we have a partial differential equation as you can tell. it'd be awfully nice for this to reduce to an ordinary differential equation, but that'd require \(u\) be a function of either \(x\) or \(y\) but not both. in general you compare the coefficients of our partial derivatives to the coefficient of \(u\) and if they cancel out to eliminate a variable we want to keep that term and annihilate the other by setting the other partial derivative to \(0\). unfortunately here is a problem in which no such nice \(u\) can be found

Answer 5

oh wait I screwed all that work above up

Answer 6

(1)$$y(4+y)\,dx-2(x^2-y)\,dy=0$$if you check you'll see this is not an exact equation:$$\frac{\partial}{\partial y}y(4+y)=4+2y,\frac{\partial}{\partial x}(-2(x^2-y))=-4x\\4+2y\ne-4x$$now imagine we had a integration factor \(u(x,y)\) that turned our equation into an exact one:$$uy(4+y)\,dx-2u(x^2-y)\,dy=0$$which gives the mixed partials:$$\frac{\partial}{\partial y}uy(4+y)=(4+2y)u+y(4+y)\frac{\partial u}{\partial y}\\\frac{\partial}{\partial x}(-2u(x^2-y))=-4xu-2(x^2-y)\frac{\partial u}{\partial x}$$since we want our equation to now be exact we equate the two:$$u(4+2y)+y(4+y)\frac{\partial u}{\partial y}=-4xu-2(x^2-y)\frac{\partial u}{\partial x}\\u(4+4x+2y)=-y(4+y)\frac{\partial u}{\partial y}-2(x^2-y)\frac{\partial u}{\partial x}$$

Answer 7

is this method applicable to # 2 and 3? o.o

Answer 8

okay. reviewing last night's lesson from you... you made it partial derivative of u over partial derivative of y (or x) = 0, hence getting one of each partial dertivative of y (or x) and.. um, making it the denominator of the other side, getting derivative of u / derivative of y ( or x)... what should I do next after that? o.o