微分方程 x(dx/dy)-y-根号(x^2+y^2)=0的通解
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发布时间：2024-03-29 22:52
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热心网友
时间：2024-08-16 21:37
(x/y)(dx/dy)-1-√((x/y)^2+1)=0
令x/y=u ，代入：
u(u+yu')=√(u^2+1)+1
yu'= (√(u^2+1)+1)/u-u= (√(u^2+1)+1-u^2)/u
udu/ (√(u^2+1)+1-u^2)=dy/y
du^2/ (√(u^2+1)+1-u^2)=2dy/y
积分∫dt/(√(t+1)+1-t)可令√(t+1)=z化成有理分式函数求解得：
∫dt/(√(t+1)+1-t)= (-2/3)ln(√(t+1)+1)+(-4/3)ln(√(t+1)-2),代入得通解：
(-2/3)ln(√(u^2+1)+1)+(-4/3)ln(√(u^2+1)-2)=2lny+(-2/3)lnC
或通解：(√(u^2+1)(√(u^2+1)-2)^2=C/y^3 其中：u=x/y
热心网友
时间：2024-08-16 21:37
题目相差仅一个负号。。。方法类似。
DIfferential Equation Homogenous help dy/dx = (y - sqrt(x^2+y^2))/ x?
The question i have is to solve the given intial value problem dy/dx = (y - sqrt(x^2+y^2))/ x. Y (3) = 4 The problem i am having is how do i even prove it homogenous? I have no idea where to go from this form please help =/
2 years ago
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The differential equation dy/dx = (y - sqrt(x^2+y^2))/x can be rewritten in the form
dy/dx = y/x - sqrt[1 + (y/x)^2].
So, let y = xv, and dy/dx = v + x dv/dx.
Then, the DE transforms to
v + x dv/dx = v - sqrt(1 + v^2)
==> dv / sqrt(1 + v^2) = -dx/x.
Integrate both sides:
ln |sqrt(1 + v^2) + v| = -ln |x| + A
-------------------------
For the left side, note that
∫ dv / sqrt(1 + v^2)
= ∫ (sec^2(t) dt) / sec t, by setting v = tan t
= ∫ sec t dt
= ln |sec t + tan t| + C
= ln |sqrt(1 + v^2) + v| + C.
--------------------------
Simplifying further:
sqrt(1 + v^2) + v = C/x, with C = e^A
==> sqrt(1 + (y/x)^2) + y/x = C/x
==> sqrt(x^2 + y^2) + y = C.
To solve for C, use y(3) = 4:
sqrt(3^2 + 4^2) + 4 = C ==> C = 9.
Thus, the desired solution is
sqrt(x^2 + y^2) + y = 9.
I hope this helps!
