Question #289441

Using Charpit’s method, solve the equation:


zp² -y²p +y²q =0

1
Expert's answer
2022-01-25T07:03:15-0500

Using the Charpit's method, we shall solve PDE

zp2y2p+y2qzp²-y²p+y²q


Consider

f(x,y,z,p,q)=0f(x,y,z,p,q)=0


Given the PDE zp2y2p+y2qzp²-y²p+y²q

We have that

f(x,y,z,p,q)f(x,y,z,p,q) =zp2y2p+y2q=0=zp²-y²p+y²q=0


We have the formula

dpfx+pfz=dqfy+qfz=dzpfpqfq=dxfp=dyfq\displaystyle\frac{dp}{\frac{\partial f}{\partial x}+p\frac{\partial f}{\partial z}}=\frac{dq}{\frac{\partial f}{\partial y}+q\frac{\partial f}{\partial z}}=\frac{dz}{-p\frac{\partial f}{\partial p}-q\frac{\partial f}{\partial q}}=\frac{dx}{-\frac{\partial f}{\partial p}}=\frac{dy}{-\frac{\partial f}{\partial q}}


It follows that

fx=0,fy=2yp+2yq,fz=p2,fp=2pz2py,fq=y2\frac{\partial f}{\partial x}=0,\frac{\partial f}{\partial y}=-2yp+2yq,\frac{\partial f}{\partial z}=p²,\frac{\partial f}{\partial p}=2pz-2py,\frac{\partial f}{\partial q}=y²


Substituting, we have

    dpp3=dq2py+2qy+p2q=dzp(2pzy2)q(y2)=dyy2\displaystyle \implies\frac{dp}{p³}=\frac{dq}{2py+2qy+p²q}=\frac{dz}{-p(2pz-y²)-q(y²)}=\frac{dy}{-y²}



Consider

dpp3=dyy2\frac{dp}{p³}=\frac{dy}{-y²}

We have that

p3dp=y2dy\displaystyle p^{-3}dp=-y^{-2}dy


Integrating, we have

    p3dp=y2dy    p3dp=y2dy\implies\int p^{-3}dp=\int -y^{-2}dy\\\implies\int p^{-3}dp=-\int y^{-2}dy


    p22=y1+c\implies \frac{p^{-2}}{-2}=y^{-1}+c

Where cc is the integrating constant.

    1p2=2y+c    y=c2p2\implies \frac{1}{p²}=\frac{-2}{y}+c\\\implies y=c-2p²

by making pp the subject of the relation

    p=cy2\implies p=\sqrt{\frac{c-y}{2}}


Now, Substituting the value of pp into the given PDE

We have

    z(ay2)y2ay2+y2q=0\implies z(\frac{a-y}{2})-y²\sqrt{\frac{a-y}{2}}+y²q=0


    y2q=y2ay2z(ay2)\implies y²q=y²\sqrt{\frac{a-y}{2}}-z(\frac{a-y}{2})


    q=ay2zy2(ay2)\implies q=\sqrt{\frac{a-y}{2}}-\frac{z}{y²}(\frac{a-y}{2})


we have been able to get values for pandqp\hspace{0.1cm}and\hspace{0.1cm}q


NOW,

we know that dz=pdx+qdydz=p\hspace{0.05cm}dx+q\hspace{0.05cm}dy

By substituting pandqindzp\hspace{0.1cm}and\hspace{0.1cm}q\hspace{0.1cm}in\hspace{0.1cm}dz


We have

dz=ay2dx+{ay2zy2(ay2)}dydz=\sqrt{\frac{a-y}{2}}dx+\left\{\sqrt{\frac{a-y}{2}}-\frac{z}{y²}(\frac{a-y}{2})\right\}dy


It follows that,


dz=ay2dx+{ay2zy2(ay2)}dy\displaystyle dz=\int\sqrt{\frac{a-y}{2}}dx+\int \left\{\sqrt{\frac{a-y}{2}}-\frac{z}{y²}(\frac{a-y}{2})\right\}dy


By integration, we have our final result to be


z=xcy2+[2(cy)3(yc)+zc+ylog(y)2y]\displaystyle z=\frac{x\sqrt{c-y}}{\sqrt{2}}+\left[\frac{\sqrt{2(c-y)}}{3}(y-c)+z\frac{c+y\hspace{0.05cm}log(y)}{2y}\right]



Which is the solution of the PDE

zp2y2p+y2qzp²-y²p+y²q



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS