## [大學]偏微分

### [大學]偏微分

grove888

grove888

grove888 寫到:抱歉,請點選"偏微分方程"

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Solve: dy/dx = 2xy/(x^2 - y^2)
[solution]
(x^2-y^2) dy = 2xy dx = y d(x^2)
Let X = x^2, then after dividing y on both sides, one can get
(X/y - y) dy =  dX
Let Y = y^2, then
y = Y^(1/2), and  dY = 2y dy = 2Y^(1/2) dy.
So dy =(1/2) *Y(-1/2) dY and then
[ X/Y^(1/2) - Y^(1/2) ] dY = 2*Y^(1/2) dX, or
(X-Y) dY = 2Y dX.
Add 2YdY on both sides, one can get
(X+Y) dY = 2Y(dX+dY) = 2Y d(X+Y).

Let Z = X+Y, then one can get separable differential equation as
Z dY = 2Y dZ, or
d(lnY) = 2 d(lnZ)
ln Y = 2 lnZ + C1
Y = BZ^2.

Rewrite the solution using the (x,y) variables:
y^2 = B(x^2 + y^2)^2
Take square root on both sides,
y = c (x^2 + y^2)
[QED]

lskuo

lskuo 寫到:
(X-Y) dY = 2Y dX.
Add 2YdY on both sides, one can get
(X+Y) dY = 2Y(dX+dY) = 2Y d(X+Y).

How did I know to add 2YdY to make it as a separable form? Consider the following problem:
dy/dx = y/(ax+by)
(ax+by) dy = ydx
Adding cydy on both sides, one can get
[ax + (b+c)y] dy = yd(x+cy)

If one chooses c such that
ax + (b+c) y = a(x+cy).
In other words,
c = b/(a-1), then the equation becomes
a(x+cy)dy = yd(x+cy), which is a separable DE.

For the case of a=1:
dy/dx = y/(x+by)
(x+by) dy = ydx , or bydy = ydx - xdy.

Recall d(x/y) = (ydx - xdy)/y^2. Thus,
b d(lny) = d(x/y)
b ln(y) = x/y + c, or
y^b = C exp(x/y).

lskuo

grove888

### Re: [大學]解偏微分!

grove888 寫到:多謝仁兄解題,想不到解此題竟如此複雜!

lskuo

grove888