Question

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

Answer 1

Solve by Monge’s method

Comparing the given equation with $Rr+Ss+Tt=V,$we have $R=1, S=a+b,T=ab,V=xy.$

Here Monge’s subsidiary equations

$$Rdpdy+Tdqdx-Vdydx=0$$

$$Rdy^2-Sdxdy+Tdx^2=0$$

become

$$dpdy+abdqdx-xydydx=0 \ \ \ \ \ \ \ \ \ \ (1)$$

$$dy^2-(a+b)dxdy+abdx^2=0 \ \ \ \ \ \ \ \ (2)$$

Equation $(2)$ may be factorised as

$$(dy-bdx)=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3)$$

$$(dy-adx)=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$$

On integration

$$y-bx=c_1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (5)$$

$$y-ax=c_2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (6)$$

Combining equation (3) with subsidiary equation (1), we get

$$dp(bdx)+abdqdx-xy(bdx)dx=0$$

$$dp+adq-xydx=0$$

By using (5)

$$dp+adq-x(c_1+bx)dx=0$$

On integration, we get

$$p+aq-(\dfrac{c_1}{2})x^2-(\dfrac{b}{3})x^3=c_3$$

By using (5)

$$p+aq-\dfrac{x^2}{2}(y-bx)-(\dfrac{b}{3})x^3=c_3$$

$$p+aq-(\dfrac{1}{2})x^2y+(\dfrac{1}{6})bx^3=c_3$$

Therefore the first intermediate integral is

$$p+aq-\dfrac{1}{2}x^2y+\dfrac{1}{6}bx^3=f_1(y-bx)\ \ \ \ \ \ \ \ (7)$$

Similarly, the second intermediate integral corresponding to equation (4) is

$$p+bq-\dfrac{1}{2}x^2y+\dfrac{1}{6}ax^3=f_2(y-ax)\ \ \ \ \ \ \ \ (8)$$

Now from above two intermediate integrals (7) and (8), we deduce the values of p and q as

$$q=\dfrac{1}{6}x^3+\dfrac{1}{a-b}\big[f_1(y-bx)-f_2(y-ax)\big]$$

$$p=\dfrac{1}{2}x^2y-\dfrac{1}{6}(a+b)x^3+\dfrac{1}{a-b}\big[af_2(y-ax)-bf_1(y-bx)\big]$$

Substituting these values of $p$ and $q$ in $dz=pdx+qdy,$ we get

$$dz=\dfrac{1}{2}x^2ydx-\dfrac{1}{6}(a+b)x^3dx+$$

$$+\dfrac{1}{a-b}\big[af_2(y-ax)dx-bf_1(y-bx)dx\big]+$$

$$+\dfrac{1}{6}x^3dy+\dfrac{1}{a-b}\big[f_1(y-bx)dy-f_2(y-ax)dy\big]$$

Or

$$dz=\dfrac{1}{6}(3x^2ydx+x^3dy)-\dfrac{1}{6}(a+b)x^3dx-$$

$$-\dfrac{1}{b-a}\big[af_2(y-ax)dx-bf_1(y-bx)dx\big]-$$

$$-\dfrac{1}{b-a}\big[f_1(y-bx)dy-f_2(y-ax)dy\big]$$

Or

$$dz=\dfrac{1}{6}d(x^3y)-\dfrac{1}{6}(a+b)x^3dx+$$

$$+\dfrac{1}{b-a}f_2(y-ax)(dy-adx)-$$

$$-\dfrac{1}{b-a}f_1(y-bx)(dy-bdx)$$

Integrating, we get the required solution as

$$z=\dfrac{1}{6}x^3y-\dfrac{1}{24}(a+b)x^4+\phi(y-ax)+\phi_1(y-bx)$$