Question

Solve the partial differential equation r+ (a+b)s +abt = xy

Answer 1

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