微分方程求解,要过程? 20
3个回答
展开全部
3.设y=tx,则dy=xdt+tdx,原方程变为
dx/x=(t^2-3)dt/(5t-t^3)=(-1/5)[3/t+1/(t-√5)+1/(t+√5)]dt,
积分得lnx=(1/5)ln[t^3*(t^2-5)]+lnc
所以x=c*[t^3*(t^2-5)]^(-1/5),
1=c[y^3*(y^2-5x^2)]^(-1/5),
y(0)=1,
所以c=1,
y^3*(y^2-5x^2)=1,为所求。
解2 原方程两边都乘以5y^2,得
(5y^4-15x^2y^2)dy-10xy^3dx=0,
即d(y^5-5x^2y^3)=0,
积分得y^5-5x^2y^3=c,
y(0)=1,
所以c=1,y^5-5x^2y^3=1,为所求。
4.设y=xc(x)是y'-y/x=(-2/x)lnx的解,
则y'=c(x)+xc'(x),
代入上式得c'(x)=(-2/x^2)lnx,
积分得c(x)=(2/x)lnx+2/x+c,
所以y=2lnx+2+cx,
y(1)=1,
所以c=-1.y=2lnx+2-x.
dx/x=(t^2-3)dt/(5t-t^3)=(-1/5)[3/t+1/(t-√5)+1/(t+√5)]dt,
积分得lnx=(1/5)ln[t^3*(t^2-5)]+lnc
所以x=c*[t^3*(t^2-5)]^(-1/5),
1=c[y^3*(y^2-5x^2)]^(-1/5),
y(0)=1,
所以c=1,
y^3*(y^2-5x^2)=1,为所求。
解2 原方程两边都乘以5y^2,得
(5y^4-15x^2y^2)dy-10xy^3dx=0,
即d(y^5-5x^2y^3)=0,
积分得y^5-5x^2y^3=c,
y(0)=1,
所以c=1,y^5-5x^2y^3=1,为所求。
4.设y=xc(x)是y'-y/x=(-2/x)lnx的解,
则y'=c(x)+xc'(x),
代入上式得c'(x)=(-2/x^2)lnx,
积分得c(x)=(2/x)lnx+2/x+c,
所以y=2lnx+2+cx,
y(1)=1,
所以c=-1.y=2lnx+2-x.
展开全部
3. y' = dy/dx = 2xy/(y^2-3x^2) = 2(y/x)/[(y/x)^2-3] 是齐次方程
令 y = xp, 则 y' = p + xdp/dx = 2p/(p^2-3)
xdp/dx = -(p^3-5p)/(p^2-3), (p^2-3)dp/(p^3-5p) = -dx/x,
(1/5)[3/p+1/(p+√5)+1/(p-√5)]dp = -dx/x
(1/5)[3ln|p|+ln|p+√5|+ln|p-√5|] = - ln|x| + (1/5)lnC
3ln|p|+ln|p+√5|+ln|p-√5|+5ln|x| = lnC
x^5 p^3 (p^2-5) = C, 即 y^3(y^2-5x^2) = C
y(0) = 1 代入, 得 C = 1, 则 y^3(y^2-5x^2) = 1.
4. 一阶线性微分方程
y = e^(∫dx/x)[∫(-2lnx/x)e^(-∫dx/x)dx + C]
= x[∫(-2lnx/x^2)dx + C] = x[∫2lnxd(1/x) + C]
= x[2lnx/x - 2∫dx/x^2 + C] = x[2lnx/x + 2/x + C] = 2lnx + 2 + Cx
y(1) = 1 代入得 C = -1
y = 2lnx - x + 2
令 y = xp, 则 y' = p + xdp/dx = 2p/(p^2-3)
xdp/dx = -(p^3-5p)/(p^2-3), (p^2-3)dp/(p^3-5p) = -dx/x,
(1/5)[3/p+1/(p+√5)+1/(p-√5)]dp = -dx/x
(1/5)[3ln|p|+ln|p+√5|+ln|p-√5|] = - ln|x| + (1/5)lnC
3ln|p|+ln|p+√5|+ln|p-√5|+5ln|x| = lnC
x^5 p^3 (p^2-5) = C, 即 y^3(y^2-5x^2) = C
y(0) = 1 代入, 得 C = 1, 则 y^3(y^2-5x^2) = 1.
4. 一阶线性微分方程
y = e^(∫dx/x)[∫(-2lnx/x)e^(-∫dx/x)dx + C]
= x[∫(-2lnx/x^2)dx + C] = x[∫2lnxd(1/x) + C]
= x[2lnx/x - 2∫dx/x^2 + C] = x[2lnx/x + 2/x + C] = 2lnx + 2 + Cx
y(1) = 1 代入得 C = -1
y = 2lnx - x + 2
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询