两道高数微分方程题,求详细过程!求高手指教啊!(y^4-3x^2)dy+xydx=0;y^3dx+2(x^2-xy^2)dy=0.
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![两道高数微分方程题,求详细过程!求高手指教啊!(y^4-3x^2)dy+xydx=0;y^3dx+2(x^2-xy^2)dy=0.](/uploads/image/z/7236146-2-6.jpg?t=%E4%B8%A4%E9%81%93%E9%AB%98%E6%95%B0%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8B%E9%A2%98%2C%E6%B1%82%E8%AF%A6%E7%BB%86%E8%BF%87%E7%A8%8B%21%E6%B1%82%E9%AB%98%E6%89%8B%E6%8C%87%E6%95%99%E5%95%8A%21%EF%BC%88y%5E4-3x%5E2%29dy%2Bxydx%3D0%3By%5E3dx%2B2%28x%5E2-xy%5E2%29dy%3D0.)
两道高数微分方程题,求详细过程!求高手指教啊!(y^4-3x^2)dy+xydx=0;y^3dx+2(x^2-xy^2)dy=0.
两道高数微分方程题,求详细过程!求高手指教啊!
(y^4-3x^2)dy+xydx=0;
y^3dx+2(x^2-xy^2)dy=0.
两道高数微分方程题,求详细过程!求高手指教啊!(y^4-3x^2)dy+xydx=0;y^3dx+2(x^2-xy^2)dy=0.
dx/dy+(y^4-3x^2)/xy=0
dx/dy+y^3/x-3x/y=0
设x/y=u,则x=yu
u+ydu/dy+y^2/u-3u=0
du/dy+y/u-2u/y=0
设u/y=t,则u=yt
t+ydt/dy+1/t-2t=0
dy/y=dt/(t-1/t)
C1+lny=ln(t^2-1)^1/2
C2y^2=y^2-1
将t,u代入得
Cy^6=x^2-y^4 (C1C2C是积分常数)
dx/dy+2(x^2-xy^2)/y^3=0
dx/dy+2x^2/y^3-2x/y=0
设x/y=u则x=yu
u+ydu/dy+2u^2/dy-u/y=0
设u/y=t则u=yt
t+ydt/dy+2t^2-t=0
dt/dy+2t^2/y=0
-dt/2t^2=dy/y
1/2t=lnY+C1 (Y为y的绝对值,C1为积分常数)
把u,t代入得
y^2/2x=lnY+C1
1。(y^4-3x²)dy+xydx=0
显然,y=0是原方程的解
于是,设x=ty² (y≠0),则dx=y²dt+2tydy
代入原方程得(y^4-3t²y^4)dy+ty³(y²dt+2tydy)=0
==>(1-t²)...
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1。(y^4-3x²)dy+xydx=0
显然,y=0是原方程的解
于是,设x=ty² (y≠0),则dx=y²dt+2tydy
代入原方程得(y^4-3t²y^4)dy+ty³(y²dt+2tydy)=0
==>(1-t²)dy+tydt=0
==>tdt/(t²-1)=dy/y
==>1/2[1/(t-1)-1/(t+1)]dt=dy/y
==>ln│(t-1)/(t+1)│=2ln│y│+ln│C│ (C是积分常数)
==>(t-1)/(t+1)=Cy²
==>(x-y²)/(x+y²)=Cy²
∴原方程的通解是y=0与(x-y²)/(x+y²)=Cy² (C是积分常数);
2。y³dx+2(x²-xy²)dy=0
显然,y=0是原方程的解
于是,设x=ty² (y≠0),则dx=y²dt+2tydy
代入原方程得y³(y²dt+2tydy)+2(t²y^4-ty^4)dy=0
==>ydt+2t²dy=0
==>2dy/y=-dt/t²
==>2ln│y│=1/t+ln│C│ (C是积分常数)
==>y²=Ce^(1/t)
==>y²=Ce^(y²/x)
∴原方程的通解是y=0与y²=Ce^(y²/x) (C是积分常数)。
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