The given differential equation r+ (a+b)s +abt = xy is of the form
Rr + Ss + Tt= v where r,s,t ,v are the functions of x and y.
This type of D.E can be solved by using MONGE'S METHOD.
Monge's subsidiary equations are
"1. R\\space dpdy+ T dqdx- v \\space dxdy = 0\\\\2. R \\space dy^2- S \\space dxdy+ T \\space dx^2=0"
Now comparing with the given equation we have
"R=1,S=(a+b),T=ab \\space and \\space v= xy"
Therefore
"dpdy+ab \\space dqdx - xy\\space dxdy=0" ...............(1)
"dy^2-(a+b)\\space dxdy+ ab\\space dx^2=0" .............(2)
From (2)
"dy^2-(a+b)\\space dxdy+ ab\\space dx^2=0\\\\or, dy^2-a \\space dxdy -b\\space dxdy+ ab\\space dx^2=0\\\\or, dy(dy-adx)-bdx(dy -adx)=0\\\\or, (dy-adx)(dy-bdx)=0\\\\\\therefore (dy-adx)=0 .........(3)\\\\ \\implies y-ax = A \\space (after \\space integration) ......(4)\\\\or, (dy-bdx)=0..........(5)\\\\ \\implies y-bx=B (after \\space integration)........(6)\\\\where A \\space and \\space B are \\space constants."
Now using equations (1) and (3) we get
"dpdy+ab \\space dqdx - xy\\space dxdy=0\\\\or, dpdy+b \\space dqdy - xy\\space dxdy=0\\\\or, dy(dp+b \\space dq-xydx)=0\\\\ As \\space dy\\neq0 \\\\\\therefore (dp+b \\space dq-xydx)=0\\\\"
Integrating we get
"\\int dp+b\\int dq- \\int x(A+ax)dx=f_1(y-ax)" (because y =A+ax from (4) )
"or, p+bq-A \\frac{x^2}{2}-a\\frac{x^3}{3}=f_1(y-ax)\\\\or, p+bq-(y-ax) \\frac{x^2}{2}-a\\frac{x^3}{3}=f_1(y-ax)\\\\or, p+bq- \\frac{x^2y}{2}+\\frac{ax^3}{6}=f_1(y-ax)" .........................(7)
Integrating we get
"\\int dp+a\\int dq- \\int x(B+bx)dx=f_2(y-bx)" (because y =B+bx from (6) )
"or, p+aq-B \\frac{x^2}{2}-b\\frac{x^3}{3}=f_2(y-bx)\\\\or, p+aq-(y-bx) \\frac{x^2}{2}-b\\frac{x^3}{3}=f_2(y-bx)\\\\or, p+aq- \\frac{x^2y}{2}+\\frac{bx^3}{6}=f_2(y-bx)" ................................(8)
Now solving equations (7) and (8) simultaneously we get
"p+bq- \\frac{x^2y}{2}+\\frac{ax^3}{6}=f_1(y-ax)\\\\p+aq- \\frac{x^2y}{2}+\\frac{bx^3}{6}=f_2(y-bx)"
solving we get
"p= \\frac{x^2y}{2}-\\frac{x^3(a+b)}{6}+\\frac{af_1(y-ax)-bf_2(y-bx)}{(a-b)} \\space and \\\\q = \\frac{x^3}{6}+\\frac{f_2(y-bx)-f_1(y-ax)}{(a-b)}"
Now solution of the Given DE is obtained by substituting p and q in
"dz =p \\space dx+q \\space dy"
"\\therefore dz = \\{\\frac{x^2y}{2}-\\frac{x^3(a+b)}{6}+\\frac{af_1(y-ax)-bf_2(y-bx)}{(a-b)} \\} \\space dx+ \\{\\frac{x^3}{6}+\\frac{f_2(y-bx)-f_1(y-ax)}{(a-b)}\\} \\space dy"
"dz = \\frac{x^2y}{2} dx-\\frac{x^3(a+b)}{6} dx+\\frac{af_1(y-ax)-bf_2(y-bx)}{(a-b)} dx+ \\frac{x^3}{6}dy+\\frac{f_2(y-bx)-f_1(y-ax)}{(a-b)}dy"
"dz = \\frac{x^2y}{2} dx + \\frac{x^3}{6}dy-\\frac{x^3(a+b)}{6} dx+\\frac{1}{(a-b)}\\{{af_1(y-ax)dx-bf_2(y-bx)dx}+f_2(y-bx)dy-f_1(y-ax)dy\\}"
"dz = \\frac{1}{6}(3x^2ydx+x^3dy)- \\frac{(a+b)}{6}x^3dx-\\frac{1}{(a-b)}\\{f_1(y-ax)*(dy-adx)-f_2(y-bx)*(dy-bdx)\\}"
"dz = \\frac{1}{6}d(x^3y) - \\frac{(a+b)}{6}x^3dx - \\frac{1}{(a-b)}\\{f_1(y-ax)*(dy-adx)-f_2(y-bx)*(dy-bdx)\\}"
Now Integrating we get
"\\int dz = \\frac{1}{6}\\int d(x^3y) - \\frac{(a+b)}{6} \\int x^3dx - \\frac{1}{(a-b)}\\{\\int f_1(y-ax)*(dy-adx)- \\int f_2(y-bx)*(dy-bdx)\\}"
"z = \\frac{x^3y}{6}-\\frac{(a+b)x^4}{24}- \\frac{1}{(a-b)}\\{\\phi_1(y-ax) -\\phi_2(y-bx)\\}"
This is the complete solution
Comments
Leave a comment