xy'-y-√y2-x2=0的通解 求微分方程y'=(y-x+1)/(y+x+5)的通解

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xy'-y-√y2-x2=0的通解 求微分方程y'=(y-x+1)/(y+x+5)的通解 y'=y/y-xxy'-y-√y2-x2=0的通解let u= y^2-x^2 u=y/x du/dx = (1/x) dy/dx -(1/x^2)y (1/x)dy/dx = du/dx + (1/x)u dy/dx =x[du/dx + (1/x)u] -------- xy'-y-√(y^2-x^2)=0 y'-(y/x)-√[(y/x)^2-1]=0 x[du/dx + (1/x)u] - u - √(u^2-1)=0 xdu/dx - √(u^2-1)=0 ∫ du/√(u^2-1) xy'-y-√y2-x2=0的通解let u= y^2-x^2 u=y/x du/dx = (1/x) dy/dx -(1/x^2)y (1/x)dy/dx = du/dx + (1/x)u dy/dx =x[du/dx + (1/x)u] -------- xy'-y-√(y^2-x^2)=0 y'-(y/x)-√[(y/x)^2-1]=0 x[du/dx + (1/x)u] - u - √(u^2-1)=0 xdu/dx - √(u^2-1)=0 ∫ du/√(u^2-1)

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求微分方程y'=(y-x+1)/(y+x+5)的通解

设x=u+a, y=v+b, a+b+5=0, b-a+1=0 => a=-2, b=-3 dv/du=(v-u)/(v+u) 设v=u p, p+u dp/du=(p-1)/(p+1) (p+1)/(p²+1) dp=-du/u 1/2 ln(1+p²)+atan p=-ln|u| +C1 p=(y+3)/(x+2), u=x+2 √(1+(y+3)²/(x+2)²) e^((y+3)/(x+2))=C/

求齐次微分方程y'=x/y+y/x满足x=1y=4的解

求齐次微分方程y'=x/y+y/x满足x=1y=4的解设z = y/x ==> dy/dx = z + x dz/dx y' = x/y + y/x z + x dz/dx = 1/z + z x dz/dx = 1/z z dz = 1/x dx z²/2 = ln|x| + c₁/2 (y/x)² = 2ln|x| + c₁ 初值条件:y|_(x=1) = 4 (4/1)² = 2ln(1) + c₁ 即 c₁

2(xy+x)y'=y 求这个微分方程的通解

2x(y+1)y'=y 2(y+1)dy/y=dx/x 2(1+1/y) dy = dx/x 2y+2ln|y| = ln|x| +ln|C| ln|x| +ln|C| - 2ln|y| =2y ln|Cx/y²| = 2y Cx/y² =e^(2y) Cx=y² e^(2y)

求微分方程 y'=y/(x+y)+(x+y)^(3/2)的通解

设p=y' 则y"=dp/dx=dp/dy* dy/dx=pdp/dy 方程化为:pdp/dy=p^3+p dp/dy=p^2+1 dp/(p^2+1)=dy arctanp=y+c p=tan(y+c) dy/dx=tan(y+c) dy/tan(y+c)=dx cos(y+c)dy/sin(y+c)=dx d(sin(y+c))/sin(y+c)=dx ln[sin(y+c)]=x+c1 sin(y+c)=c2e^x

求微分方程y''+(1/x)y'=(1/x)的通解

xy''+y'=1 (xy')'=1 xy'=x+C y'=1+C/x y=x+Clnx+D

请帮忙写出下列方程的特解形式,y''-y+y=x^2e^x

继续求导数 y'''+y"=2x y""+y'''=2 y""=-(y'''-2) y"""/(y'''-2)=-1 [ln(y'''-2)]'=-1 ln(y'''-2)=-x+C1 y'''-2=C2e^(-x) y'''=C2e^(-x)+2 y"=-C2e^(-x)+2x+C3 y'=C2e^(-x)+x²+C3x+C4 y=-C2e^(-x)+x³/3+C3x²/2+C4x+C

高等数学二阶方程初值问题 求解二阶方程2yy''=(y')...

令y'=p,y''=p dp/dy 原式转换dp/dy=p/2y+y/2p 用伯努力方程得到(z=p¹-ⁿ=p²)dz/dy-z/y=y 一阶线性非齐次方程通解z=e-∫-1/ydy﹝C+∫ye∫-1/ydy﹞ p=根号(Cy +y²) y=∫根号(Cy +y²)dx y²=(Cy +y²)x²

xy'-y-√y2-x2=0的通解

xy'-y-√y2-x2=0的通解let u= y^2-x^2 u=y/x du/dx = (1/x) dy/dx -(1/x^2)y (1/x)dy/dx = du/dx + (1/x)u dy/dx =x[du/dx + (1/x)u] -------- xy'-y-√(y^2-x^2)=0 y'-(y/x)-√[(y/x)^2-1]=0 x[du/dx + (1/x)u] - u - √(u^2-1)=0 xdu/dx - √(u^2-1)=0 ∫ du/√(u^2-1)

y''-a(y')∧2=0,y|(x=0)=0,y'|(x=0)=-1,求特解

y''/y'^2=a (1/y')'=-a 1/y'=-ax+C1 因为y'(0)=-1,-1=C1,所以1/y'=-ax-1 y'=-1/(ax+1) y=(-1/a)*ln|ax+1|+C2 因为y(0)=0,0=C2,所以y=(-1/a)*ln|ax+1|

标签: y'=y/y-x xy'-y-√y2-x2=0的通解

回答对《求微分方程y'=(y-x+1)/(y+x+5)的通解》的提问

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