Question #269452

Homogeneous differential equation


1. (xy^2)dx-(x^3+y^3)dy=0


2. (x^2+y^2)dx+xydy=0


3. (y^2-x^2)dx+2xydy=0


4. (3x+2y)dx-2xdy=0

1
Expert's answer
2021-11-22T15:09:19-0500

1.


(xy2)dy(x3+y3)dx=0(xy^2)dy-(x^3+y^3)dx=0

(xy2)y(x3+y3)=0(xy^2)y'-(x^3+y^3)=0

y1xy=x2y2y'-\dfrac{1}{x}y=x^2y^{-2}

First order Bernoulli ODE


u=y1(2)=y3u=y^{1-(-2)}=y^3

u=3y2yu'=3y^2y'

13u1xu=x2\dfrac{1}{3}u'-\dfrac{1}{x}u=x^2

u3xu=3x2u'-\dfrac{3}{x}u=3x^2

Integration factor


μ(x)=e(3/x)dx=x3\mu(x)=e^{\int(-3/x)dx}=x^{-3}

x3u3x4u=3xx^{-3}u'-\dfrac{3}{x^4}u=\dfrac{3}{x}d(x3u)=3xdxd(x^{-3}u)=\dfrac{3}{x}dx

Integrate


d(x3u)=3xdx\int d(x^{-3}u)=\int\dfrac{3}{x}dx

x3u=lnx+Cx^{-3}u=\ln x+C

y3=3x3lnx+Cx3y^3=3x^3\ln x+Cx^3

y=x3lnx+C3y=x\sqrt[3]{3\ln x+C}

2.


(x2+y2)dx+xydy=0(x^2+y^2)dx+xydy=0

x2+y2+xyy=0x^2+y^2+xyy'=0

y+1xy=xy1y'+\dfrac{1}{x}y=-xy^{-1}

First order Bernoulli ODE


u=y1(1)=y2u=y^{1-(-1)}=y^2

u=2yyu'=2yy'

12u+1xu=x\dfrac{1}{2}u'+\dfrac{1}{x}u=-x

u+2xu=2xu'+\dfrac{2}{x}u=-2x

Integration factor


μ(x)=e(2/x)dx=x2\mu(x)=e^{\int(2/x)dx}=x^2

x2u+2xu=2x3x^2u'+2xu=-2x^3

d(x2u)=2x3dxd(x^2u)=-2x^3dx

Integrate


d(x2u)=2x3dx\int d(x^2u)=-\int2x^3dx

x2u=12x4+Cx^2u=-\dfrac{1}{2}x^4+C

y2=12x2+Cx2y^2=-\dfrac{1}{2}x^2+\dfrac{C}{x^2}

3.


(y2x2)dx+2xydy=0(y^2-x^2)dx+2xydy=0

y2x2+2xyy=0y^2-x^2+2xyy'=0

y+12xy=12xy1y'+\dfrac{1}{2x}y=\dfrac{1}{2}xy^{-1}

First order Bernoulli ODE


u=y1(1)=y2u=y^{1-(-1)}=y^2

u=2yyu'=2yy'

12u+12xu=12x\dfrac{1}{2}u'+\dfrac{1}{2x}u=\dfrac{1}{2}x

u+1xu=xy1u'+\dfrac{1}{x}u=xy^{-1}

Integration factor


μ(x)=e(1/x)dx=x\mu(x)=e^{\int(1/x)dx}=x

xu+u=x2xu'+u=x^2

d(xu)=x2dxd(xu)=x^2dx

Integrate


d(xu)=x2dx\int d(xu)=\int x^2dx

xu=13x3+Cxu=\dfrac{1}{3}x^3+C

y2=13x2+Cxy^2=\dfrac{1}{3}x^2+\dfrac{C}{x}


4.


(3x+2y)dx2xdy=0(3x+2y)dx-2xdy=0

y1xy=32y'-\dfrac{1}{x}y=\dfrac{3}{2}

Integration factor


μ(x)=e(1/x)dx=1x\mu(x)=e^{-\int(1/x)dx}=\dfrac{1}{x}

1xy1x2y=32x\dfrac{1}{x}y'-\dfrac{1}{x^2}y=\dfrac{3}{2x}

d(1xy)=32xdxd(\dfrac{1}{x}y)=\dfrac{3}{2x}dx

Integrate


d(1xy)=32xdx\int d(\dfrac{1}{x}y)=\int\dfrac{3}{2x}dx

1xy=32lnx+C\dfrac{1}{x}y=\dfrac{3}{2}\ln x+C


y=32xlnx+Cxy=\dfrac{3}{2}x\ln x+Cx



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