xy * y' = y^2 - x^2
I can't figure out why you can derive this answer,
but I just provide my method to you.
I think that this problem should multiply a integrating factor
to obtain a exact ODE, not linear ODE.
I use the traditional method the general
advanced engineering mathematics book provide.
This ODE can be rearranged to the form xydy=(y^2-x^2)dx
--> (y^2-x^2)dx-xydy=0
change into exact form Mdx+Ndy=0
multiply a integrating factor μ
--> (y^2-x^2)μdx-xyμdy=0
∫f(x)dx
use the formula μ=e
dM/dy-dN/dx
f(x)=_____________ d:partial differentiation
N
-->μ=x^(-3)
--> this ODE becomes a exact DE
-3 2 -1 -2
(x y -x )dx-(x y)dy=0
ψ=C is the solution
-3 2 -1
M= dψ/dx = x y -x ψ=y^2*x^(-2)*(-1/2) +f(y)
-2 -2
N=dψ/dy = -yx +f'(y) =-x y
so f'(y)=0 f(y)=constant
2 -2
ψ=C = y x *(-1/2)+ C1 --> y = x *( - 2ln(x) + C )^(1/2) #
↑
※ 引述《GoHarmonic (發出諧和波)》之銘言：
: xy * y' = y^2 - x^2
: 這題如果單純用變數變換來解的話，一下子就可以解出來
: 解大概是 y = x *( 2ln(x) + C )^(1/2)
: 可是我是想說，這題不知道能不能用線性的方式來解，
: 就是找Ｉ積分因子的那個方法弄成線性來解
: 我解完之後，得到的答案跟前一個答案差很多
: 1
: 得到的是 y = CX * e^(-－X^3 + XY^2)
: 3
: 想請問大家，我這樣解出來的答案，對還是不對呢？
: 謝謝
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※ 編輯: Ertkkpoo 來自: 219.70.199.84 (10/02 01:56)