Question

求解一道常微分方程题

dy/dx=(y^6-2x^2)/(2xy^5+x^2y^2)

Answer 1

解：设t=x/y³，则dx=y³dt+3ty²dy
代入原方程得dy/(y³dt+3ty²dy)=(y^6-2t²y^6)/(2ty^8+t²y^8)
==>(y³dt+3ty²dy)/dy=ty²(2+t)/(1-2t²)
==>y³dt/dy+3ty²=ty²(2+t)/(1-2t²)
==>y³dt/dy=ty²(2+t)/(1-2t²)-3ty²
==>y³dt/dy=ty²(2t+1)(3t-1)/(1-2t²)
==>(1-2t²)dt/[t(2t+1)(3t-1)]=dy/y
==>[(2/5)/(2t+1)+(7/5)/(3t-1)-1/t]dt=dy/y
==>(1/5)ln│2t+1│+(7/15)ln│3t-1│-ln│t│=ln│y│+(1/15)ln│C│ (C是积分常数)
==>ln│[(2t+1)^3][(3t-1)^7]│=ln│C(ty)^15│
==>[(2x/y³+1)^3][(3x/y³-1)^7]=C(x/y²)^15
==>[(2x+y³)^3][(3x-y³)^7]=Cx^15
故原微分方程的通解是[(2x+y³)^3][(3x-y³)^7]=Cx^15 (C是积分常数)。

Answer 2

你初值未有给定哦
function dydx=f(x,y)
dydx=(y^6-2*x^2)/(2*x*y^5+x^2*y^2);
[x,y]=ode45('f',[-30,30],2);
plot(x,y)
hold on