Question #112040
Solve charpit method: 2z+p^2+qy+2y^2=0
1
Expert's answer
2020-04-29T18:30:02-0400

Let f(x,y,z,p,q)2z+p2+qy+2y2=0f(x,y,z,p,q) \equiv 2z+p^2+qy+2y^2=0 ............(1)


Charpit's auxiliary equation are



dpfx+pfz=dqfy+qfz=dzpfpqfq=dxfp=dyfq\frac{dp}{f_x+pf_z}=\frac{dq}{f_y+qf_z}=\frac{dz}{-pf_p-qf_q}=\frac{dx}{-f_p}=\frac{dy}{-f_q}


....................(2)

From (1), fx=0, fy=q+4y, fz=2, fp=2p, fq=yf_x=0, \ f_y=q+4y, \ f_z=2, \ f_p=2p,\ f_q=y .......(3)

Using (3),(2) reduce to



dp2p=dq3q+4y=dz2p2qy=dx2p=dyy\frac{dp}{2p}=\frac{dq}{3q+4y}=\frac{dz}{-2p^2-qy}=\frac{dx}{-2p}=\frac{dy}{-y}


Taking 1st and 4th fraction we get,


    dp2p=dx2p\implies \frac{dp}{2p}=\frac{dx}{-2p}



So that , P=xP=-x ..............(4)

Now taking 2nd and 5th fraction ,we get


dq3q+4y=dyy\frac{dq}{3q+4y}=\frac{dy}{-y}




    dqdy=3q+4yy\implies \frac{dq}{dy}=-\frac{3q+4y}{y}



    dqdy+3yq=4\implies \frac{dq}{dy}+\frac{3}{y}q=-4


Which is a linear equation in first order.


I.F=e3ydy=e3logy=elogy3=y3\therefore I.F=e^{\int \frac{3}{y} dy } =e^{3logy}=e^{log {y^3} }=y^3


Now , q y3=(4)y3dy=y4q \ y^3=\int (-4)y^3dy =-y^4

    q=y\implies q=-y .


Now putting the value of p and q in

dz=pdx+qdydz=pdx+qdy we get dz=xdxydydz=-xdx-ydy

    2z=x2y2+c\implies 2z=-x^2-y^2+c

Where "c" is a integration constant.



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