Answer to Question #181336 in Differential Equations for kamal

Question #181336

solve dx/2x(y+z^2) = dy/y(2y+z^2) = dz/z^3


1
Expert's answer
2021-04-29T18:11:29-0400

The correct form of the given equation is:



dx2x(y+z2)=dy2y(y+z2)=dzz3\frac{dx}{2x(y+z^2)}=\frac{dy}{2y(y+z^2)}=\frac{dz}{z^3}

Taking the first two terms


dx2x(y+z2)=dy2y(y+z2)    dxx=dyy\frac{dx}{2x(y+z^2)}=\frac{dy}{2y(y+z^2)}\\ \implies \frac{dx}{x}=\frac{dy}{y}

Integrating both the sides, we have:


logx=logy+logc1x=yc1c1=xy(1)\log x=\log y + \log c_1\\ x=yc_1\\ c_1 = \frac{x}{y} \qquad \cdots (1)

Taking Last two terms, we have:


dx2y(y+z2)=dzz3    z3dy2y(y+z2)dz=0\frac{dx}{2y(y+z^2)}=\frac{dz}{z^3}\\ \implies z^3dy-2y(y+z^2)dz = 0

Divide both sides by z3y2-z^3y^2

1y2dydz+2zy=2z3-\frac{1}{y^2}\frac{dy}{dz}+\frac{2}{zy}=-\frac{2}{z^3}\\

Let

1y=t    1y2dydz=dtdz\frac{1}{y} =t \\ \implies -\frac{1}{y^2}\frac{dy}{dz} =\frac{dt}{dz}

On comparing it with Linear differential equation, we get:


P=2z,Q=2z3P= \frac{2}{z} ,Q= \frac{-2}{z^3}


Integrating factor, I.F.=eP.dz=e2zdz=e2logz=z2e^{\int P.dz}=e^{\int \tfrac{2}{z}dz}=e^{2logz}=z^2


The solution is:


t×I.F.=I.F.×Qdzt\times I.F.=\int I.F.\times Q dz

t(z2)=z2×2z3dz⇒t(z^2 )=∫z^2×\frac{-2}{z^3}dz

t×z2=2logz+c2\Rightarrow t\times z^2=-2logz+c_2\\ ​

c2=z2y+2logz⇒c_2= \frac{z^2}{y}+2logz

The solution of the given equation is:


ϕ(c1,c2)=0    ϕ(xy,z2y+2logz)=0\phi(c_1,c_2)=0 \implies \phi(\dfrac{x}{y},\dfrac{z^2}{y}+2logz)=0


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