Question #92510
Kindly assist me solving the following questions
1. (x^2y^2+xy)ydx + (x^2y^2-1)xdy=0
2. (x^2+y^2)dx + xdy=0
3. dy/dx +y=3y^3e^x
1
Expert's answer
2019-08-13T11:30:19-0400

1.


xy2(xy+1)dx+x(xy1)(xy+1)dy=0xy^2(xy+1)dx+x(xy-1)(xy+1)dy=0

x=0x=0y=0y=0xy+1=0=>y=1xxy+1=0=>y=-{1 \over x}

Or


y2dx+(xy1)dy=0y^2dx+(xy-1)dy=0

Using an Integrating Factor

P(x,y)=y2,Q(x,y)=xy1P(x,y)=y^2, Q(x,y)=xy-1

ðPðy=2y,ðQðx=y,ðPðyðQðx{\eth P \over \eth y}=2y, {\eth Q \over \eth x}=y, {\eth P \over \eth y}\not={\eth Q \over \eth x}ðPðyðQðx=2yy=y{\eth P \over \eth y}-{\eth Q \over \eth x}=2y-y=y

Integrating Factor

μ=μ(y)\mu=\mu(y)

1μdμdy=1P(ðPðyðQðx){1 \over \mu}{d\mu \over dy}=-{1 \over P}({\eth P \over \eth y}-{\eth Q \over \eth x})

1μdμdy=1y2(2yy){1 \over \mu}{d\mu \over dy}=-{1 \over y^2}(2y-y)

dμμ=dyy{d\mu \over \mu}=-{dy \over y}

dμμ=(dyy)=>lnμ=ln1y+lnC\int {d\mu \over \mu}=\int (-{dy \over y})=>\ln|\mu|=\ln|{1 \over y}|+\ln C

We choose


μ=1y\mu={1 \over y}

Multiplying the original differential equation by μ=1/y\mu=1/y  produces the exact equation:


ydx+(x1y)dy=0ydx+(x-{1 \over y})dy=0

Indeed, now we have


ðPðy=1,ðQðx=1,ðPðy=ðQðx{\eth P \over \eth y}=1, {\eth Q \over \eth x}=1, {\eth P \over \eth y}={\eth Q \over \eth x}

Solve the resulting equation. The function u(x,y)u(x,y) can be found from the system of equations:


{ðuðx=yðuðy=x1y\begin{cases} {\eth u \over \eth x}=y \\ {\eth u \over \eth y}=x-{1 \over y} \end{cases}

It follows from the first equation that


u=ydx=xy+φ(y)u=\int y dx=xy+\varphi (y)

Substitute this in the second equation to determine φ:\varphi:


ðuðy=x+dφdy=x1y{\eth u \over \eth y}=x+ {d \varphi \over d y}=x-{1 \over y}

dφ=(1y)dy=>φ(y)=ln(1y)\int d\varphi=\int (-{1 \over y})dy=>\varphi (y)=\ln({1 \over |y|})

Thus, the original differential equation has the following solutions:


x=0x=0y=0y=0y=1xy=-{1 \over x}xy+ln(1y)=C1xy+\ln({1 \over |y|})=C_1

where C1C_1 is a constant.


2.


(x2+y2y)dx+xdy=0(x^2+y^2-y)dx+xdy=0

We can make sure that this equation is not exact:


P(x,y)=x2+y2y,Q(x,y)=xP(x,y)=x^2+y^2-y, Q(x,y)=x

ðPðy=2y1,ðQðx=1,ðPðyðQðx{\eth P \over \eth y}=2y-1, {\eth Q \over \eth x}=1, {\eth P \over \eth y}\not={\eth Q \over \eth x}

The difference of the partial derivatives is


ðPðyðQðx=2y11=2y2{\eth P \over \eth y}-{\eth Q \over \eth x}=2y-1-1=2y-2

Using the integrating factor μ=x2+y2\mu=x^2+y^2 we find that


ðzðy=2y,ðzðx=2x{\eth z \over \eth y}=2y, {\eth z \over \eth x}=2x

Calculate the following expression:

QðzðxPðzðy=x(2x)(x2+y2y)(2y)=Q{\eth z \over \eth x}-P {\eth z \over \eth y}=x(2x)-(x^2+y^2-y)(2y)==2x2+2y22y(x2+y2)=(x2+y2)(22y)=2x^2+2y^2-2y(x^2+y^2)=(x^2+y^2)(2-2y)

As a result, we obtain the differential equation for the function μ(z):\mu(z):

1μdμdz=1(QðzðxPðzðy)(ðPðyðQðx)={1 \over \mu}{d\mu \over dz}={1 \over (Q{\eth z \over \eth x}-P {\eth z \over \eth y})}({\eth P \over \eth y}-{\eth Q \over \eth x})=

=1(x2+y2)(22y)(2y2)=1x2+y2=1z={1 \over (x^2+y^2)(2-2y)}(2y-2)=-{1 \over {x^2+y^2}}=-{1 \over z}

By integrating we get the function μ(z):\mu(z):

dμμ=(dzz)=>lnμ=lnz+lnC\int {d\mu \over \mu}=\int (-{dz \over z})=>\ln|\mu|=-\ln|z|+\ln C

We choose

μ=1z\mu={1 \over z}

We can choose the integrating factor

μ=1z=1x2+y2\mu=-{1 \over z}=-{1 \over {x^2+y^2}}

1x2+y2(x2+y2y)dx1x2+y2xdy=0-{1 \over {x^2+y^2}}(x^2+y^2-y)dx-{1 \over {x^2+y^2}}xdy=0

Or

(1yx2+y2)dx+xx2+y2dy=0(1-{y\over {x^2+y^2}})dx+{x \over {x^2+y^2}}dy=0

The general solution u(x,y)=Cu(x,y)=C is defined by the following system of equations:

{ðuðx=1yx2+y2ðuðy=xx2+y2\begin{cases} {\eth u \over \eth x}=1-{y\over {x^2+y^2}} \\ {\eth u \over \eth y}={x \over {x^2+y^2}} \end{cases}

Integrating the first equation with respect to x produces:

u(x,y)=(1yx2+y2)dx=xy1x2+y2dx=u(x,y)=\int{(1-{y\over {x^2+y^2}})}dx=x-y\int{{1\over {x^2+y^2}}}dx=

=xarctan(xy)+φ(y)=x-\arctan({x \over y})+\varphi(y)

Substituting this in the second equation, we have:


ðuðy=xx2+y2+dφdy=xx2+y2{\eth u \over \eth y}={x\over {x^2+y^2}}+ {d \varphi \over d y}={x\over {x^2+y^2}}

dφdy=0{d \varphi \over d y}=0

Thus, the general solution of differential equation in implicit form is defined by the formula:

xarctan(xy)=Cx-\arctan({x \over y})=C

where CC is a constant.



3.


dydx+y=3y3ex{dy \over dx}+y=3y^3e^xy=0y=0

Or


y3dydx+y2=3exy^{-3}{dy \over dx}+y^{-2}=3e^x

The associated homogeneous differential equation is


y3dydx+y2=0y^{-3}{dy \over dx}+y^{-2}=0

dydx+y=0{dy \over dx}+y=0

dyy=dx{dy \over y}=-dxy=C1exy=C_1e^{-x}

dydx=dC1dxexC1ex{dy \over dx}={dC_1 \over dx}e^{-x}-C_1e^{-x}

Substitute


dC1dxexC1ex+C1ex=3C13e3xex{dC_1 \over dx}e^{-x}-C_1e^{-x}+C_1e^{-x}=3C_1^3e^{-3x}e^x

dC1dx=3C13ex{dC_1 \over dx}=3C_1^3e^{-x}

dC1C13=3exdx{dC_1 \over C_1^3}=3e^{-x}dx

dC1C13=3exdx\int {dC_1 \over C_1^3}=\int3e^{-x}dx

12C12=3exdxC22-{1 \over2 C_1^2}=-3e^{-x}dx-{C_2 \over2}

C1=16exdx+C2C_1={1\over \sqrt{6e^{-x}dx+C_2}}

Thus, the original differential equation has the following solutions:


y=0y=0y=1ex6exdx+C2y={1\over e^x \sqrt{6e^{-x}dx+C_2}}

where C1C_1 is a constant.



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