CHARPIT’S METHOD: Charpit’s method is a general method for finding the complete solution of non- linear partial differential equation of the first order of the form ( ) 0 q , p , z , y , x f = . (i) Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . (ii) Integrating (ii), we get the complete solution of (i). Note: In order to integrate (ii), we must know p and q in terms of x, y, z. For this purpose, introduce another non-linear partial differential equation of the first order of the form ( ) 0 a , q , p , z , y , x F = , (iii) involving an arbitrary constant ‘a’ compatible with (i). Solving (i) and (iii), we get ( ) a , z , y , x p p = , ( ) b , z , y , x q q = . (iv) On substitution of (iv) in (ii), equation (ii) becomes integrable, resulting in the complete solution of (i) in the form 5 55 5 th th th th Topic Topic Topic Topic Partial Differential Equations Partial Differential Equations Partial Differential Equations Partial Differential Equations Method for finding the complete integral of a non- linear partial differential equation (Charpit’s Method) Prepared by: Dr. Sunil NIT Hamirpur (HP) (Last updated on 13-09-2007) Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 2 ( ) 0 b , a , z , y , x = φ , (v) containing two arbitrary constants a and b. To determine F: We differentiate (i) and (iii) partially w. r. t. x and y. Thus 0 x q . q f x p . p f p . z f x f = ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ + ∂ ∂ , (vi) 0 x q . q F x p . p F p . z F x F = ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ + ∂ ∂ , (vii) 0 y q . q f y p . p f q . z f y f = ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ + ∂ ∂ , (viii) 0 y q . q F y p . p F q . z F y F = ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ + ∂ ∂ . (ix) Eliminating x p ∂ ∂ between (vi) and (vii), we get 0 x q p f . q F p F . q f p . p f . z F p F . z f p f . x F p F . x f = ∂ ∂ | | ¹ | \ | ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ + | | ¹ | \ | ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ + | | ¹ | \ | ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ . (x) Eliminating y q ∂ ∂ between (viii) and (ix), we get 0 y p q f . p F q F . p f q . q f . z F q F . z f q f . y F q F . y f = ∂ ∂ | | ¹ | \ | ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ + | | ¹ | \ | ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ + | | ¹ | \ | ∂ ∂ ∂ ∂ − ∂ ∂ ∂ ∂ . (xi) Since y p x y z y x z x q 2 2 ∂ ∂ = ∂ ∂ ∂ = ∂ ∂ ∂ = ∂ ∂ and the last term in (x) and (xi) differ in sign only, then adding (x) and (xi), we get 0 y F q f x F p f z F q f q p f p q F z f q y f p F z f p x f = ∂ ∂ | | ¹ | \ | ∂ ∂ − + ∂ ∂ | | ¹ | \ | ∂ ∂ − + ∂ ∂ | | ¹ | \ | ∂ ∂ − ∂ ∂ − + ∂ ∂ | | ¹ | \ | ∂ ∂ + ∂ ∂ + ∂ ∂ | ¹ | \ | ∂ ∂ + ∂ ∂ , (xii) which is the linear partial differential equation (Lagrange’s linear equation) of the first order with x, y, z, p, q as independent variables and F as the dependent variable. ∴ The auxiliary equations of (xii) are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ . (xiii) These equations (xiii) are known as Charpit’s equations. Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 3 Solving (xiii), we get relations (iv) of p and q, using which, the equation (ii) is integrated resulting in the complete solution (v). Note: All the equations of Charpit’s equations (xiii) need NOT be used. Choose the simplest of (xiii), so that p and q are easily obtained. Now let us solve complete solution of non-linear partial differential equation of the first order by Charpit’s method: Q.No.1.: Solve the following non-linear partial differential equation by Charpit’s method: 0 pq qxy 2 px zx 2 2 = + − − . Sol.: Given non-linear partial differential equation is 0 pq qxy 2 px zx 2 f 2 = + − − = . (i) qy 2 px 2 z 2 x f − − = ∂ ∂ ∴ , qx 2 y f − = ∂ ∂ , x 2 z f = ∂ ∂ , q x p f 2 + − = ∂ ∂ , p xy 2 q f + − = ∂ ∂ . Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ . p xy 2 dy q x dx qxy 2 pq 2 px dz 0 dq qy 2 z 2 dp 2 2 − = − = + − = = − ⇒ . From second member, we get q = a. Putting q = a in (i), we get ( ) a x ay z x 2 p 2 − − = . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . ( ) ady dx a x ay z x 2 qdy pdx dz 2 + − − = + = ∴ dx a x x 2 ay z ady dz 2 − = − − ⇒ . Integrating on both sides, we get ( ) ( ) ( ) a x b ay z b log a x log ay z log 2 2 − = − ⇒ + − = − . ( ) a x b ay z 2 − + = ⇒ , Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 4 which is the required complete solution involving two arbitrary constants a and b. Q.No.2.: Solve the following non-linear partial differential equation by Charpit’s method: ( ) qz y q p 2 2 = + or 0 y q y p qz 2 2 = − − . Sol.: Given non-linear partial differential equation is ( ) 0 qz y q p f 2 2 = − + = . (i) 0 x f = ∂ ∂ ∴ , 2 2 q p y f + = ∂ ∂ , q z f − = ∂ ∂ , py 2 p f = ∂ ∂ , z qy 2 q f − = ∂ ∂ . Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ . z qy 2 dy py 2 dx qz dz p dq pq dp 2 + − = − = − = = − ⇒ . From the first two members, we get 0 qdy pdx = + . Integrating, we get 2 2 2 a q p = + 2 2 q a p − = ⇒ . (ii) Putting 2 2 2 a q p = + in (i), we get z y a q 2 = . ∴From (ii), we get 2 2 2 2 2 4 2 2 2 y a z z a z y a a q a p − = − = − = . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . dy z y a dx y a z z a qdy pdx dz 2 2 2 2 + − = + = ∴ . dx y a z a ydy a zdz 2 2 2 2 − = − ⇒ ( ) adx y a z y a z d 2 1 2 2 2 2 2 2 = − − ⇒ . Integrating on both sides, we get b ax y a z 2 2 2 + = − ( ) 2 2 2 2 y a b ax z + + = ⇒ , which is the required complete solution involving two arbitrary constants a and b. Q.No.3.: Solve the following non-linear partial differential equation by Charpit’s Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 5 method: 0 y 2 qy p z 2 2 2 = + + + . Sol.: Given non-linear partial differential equation is 0 y 2 qy p z 2 f 2 2 = + + + = . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ . ( ) y dy p 2 dx qy p 2 dz q 3 y 4 dq p 2 dp 2 − = − = + − = + = ⇒ . From first and fourth ratios, we get a x p dx dp + − = ⇒ − = . Substituting x a p − = in (i), we get ( ) [ ] 2 2 x a y 2 z 2 y 1 q − − − − = . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . ( ) ( ) [ ]dy x a y 2 z 2 y 1 dx x a qdy pdx dz 2 2 − + + − − = + = ∴ . Multiplying both sides by 2 y 2 , we get ( ) ( ) dy x a y 2 dy y 4 dx x a y 2 yzdy 4 dz y 2 2 3 2 2 − − − − = + Integrating on both sides, we get ( ) [ ] ( ) [ ] b y z 2 a x y b y x a y zy 2 2 2 2 4 2 2 2 = + + − ⇒ + + − − = , which is the required complete solution involving two arbitrary constants a and b. Q.No.4.: Solve the following non-linear partial differential equation by Charpit’s method: y q x p z 2 2 + = . Sol.: Given non-linear partial differential equation is 0 z y q x p f 2 2 = − + = . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ . Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 6 ( ) pq 2 dy px 2 dx y q x p 2 dz q q dq p p dp 2 2 2 2 − = − = + − = + − = + − ⇒ . From which, we have y q qydq 2 dy q x p pxdp 2 dx p 2 2 2 2 + = + . Integrating on both sides, we get ( ) ( ) a log y q log x p log 2 2 + = y aq x p 2 2 = ⇒ . (ii) From (i) and (ii), we have z y q y aq 2 2 = + ( ) 2 / 1 y a 1 z q ( ¸ ( ¸ + = ⇒ . From (ii), we have ( ) 2 / 1 x a 1 az p ( ¸ ( ¸ + = . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . ( ) ( ) dy y a 1 z dx x a 1 az qdy pdx dz 2 / 1 2 / 1 ( ¸ ( ¸ + + ( ¸ ( ¸ + = + = ∴ ( ) y dy x dx a z dz a 1 + = + ⇒ . Integrating on both sides, we get ( ) { } ( ) b y ax z a 1 + + = + [ ] ( ) a 1 b y ax z 2 + + + = ⇒ , Ans. which is the required complete solution involving two arbitrary constants a and b. Q.No.5.: Solve the following non-linear partial differential equations by Charpit’s method: yz qy pq pxy = + + . Sol. Given non-linear partial differential equation is 0 yz qy pq pxy f = − + + = . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ . ( ) ( ) ( ) ( ) ( ) ( ) y p dy q xy dx y p q q xy p dz qp q px dq y p py dp + − = + − = + − + − = + + = − + ⇒ . ( ) ( ) ( ) ( ) ( ) y p dy q xy dx y p q q xy p dz qp q px dq 0 dp + − = + − = + − + − = + + = ⇒ . Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 7 From first member, we get 0 dp = a p = ⇒ . Putting a p = in (i), we get yz qy aq axy = + + ( ) ( ) ax z y y a q − = + ⇒ ( ) y a ax z y q + − = ⇒ . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . ( ) dy y a ax z y adx qdy pdx dz + − + = + = ∴ y a ydy ax z adx dz + = − − ⇒ dy y a a 1 ax z adx dz | | ¹ | \ | + − = − − ⇒ . Integrating on both sides, we get ( ) ( ) b y a log a y ax z log + + − = − , Ans. which is the required complete solution involving two arbitrary constants a and b. Q.No.6.: Solve the following non-linear partial differential equations by Charpit’s method: pqxy z 2 = . Sol.: Given non-linear partial differential equation is 0 pqxy z f 2 = − = . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ ( ) ( ) q p q p z y z x df dy df dx qf pf dz qf f dq pf f dp = = + = + − = + − ⇒ ( ) ( ) pxy dy qxy dx pqxy 2 dz qz 2 pqx dq pz 2 pqy dp − = − = − = + − − = + − − ⇒ Using the multipliers p, q, o, x, y, we have yqz 2 ypqx qpxy ydq qdy pxz 2 xpqy pqxy xdp pdx − + − + = − + − + yqz 2 ydq qdy xpz 2 xdp pdx − + = − + ⇒ ) yq ( ) yq ( d ) xp ( ) xp ( d = ⇒ . Integrating on both sides, we get yq a xp = ay xp q = ⇒ . Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 8 Substituting ay xp q = in (i) i.e. pqxy z 2 = , we get a x p xy ay xp . p z 2 2 2 = | | ¹ | \ | = x z . a p = ⇒ . Then y a z x z . a . ay x ay xp q = = = . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . dy y z a 1 dx x z a qdy pdx dz + = + = ∴ y dy a 1 x dx a z dz + = ⇒ . Integrating on both sides, we get b / 1 b y ax z = , Ans. which is the required complete solution involving two arbitrary constants a and b. Q.No.7.: Solve the following non-linear partial differential equations by Charpit’s method: ( ) 2 yp qy px z 2 = + + . Ans.: Given non-linear partial differential equation is ( ) 0 yp qy px z 2 f 2 = − + + = . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ . ( ) ( ) y 2 dy yp 2 x 2 dx qy 2 yp 2 xp 2 dz q 2 p q 2 dq p 2 p 2 dp 2 2 = − = + − = + − − = + − ⇒ y dy yp x dx yq yp xp dz 2 p q 2 dq p 2 dp 2 2 = − = + − = | | ¹ | \ | − − = − ⇒ Using first and fifth members, we have p 2 dp y dy − = 2 2 y a ay p = = ⇒ − . Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 9 Substituting the value of p (i) i.e. ( ) 2 yp qy px z 2 = + + , we obtain | | ¹ | \ | − − | | ¹ | \ | = 2 2 2 y a x 2 z 2 y a y yq 2 3 4 2 y ax y z y 2 a q − − = ⇒ . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . dy y ax y z y 2 a dx y a qdy pdx dz 3 4 2 2 | | ¹ | \ | − − + = + = ∴ . Regrouping the terms, we get dy y 2 a y axdy aydx y zdy ydz 4 2 3 + | | ¹ | \ | − = | | ¹ | \ | + . Multiplying throughout by y, we obtain ( ) 3 2 y dy 2 a y x ad yz d + | | ¹ | \ | = . Integrating on both sides, we get b y 2 1 . 2 a y x a yz 2 2 + | | ¹ | \ | − + = . y b y 4 a y ax z 3 2 2 + − = ⇒ , Ans. which is the required complete solution involving two arbitrary constants a and b. Q.No.8.: Solve the following non-linear partial differential equations by Charpit’s method: pq qy px = + . Ans.: Given non-linear partial differential equation is 0 pq qy px f = − + ≡ . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ ( ) ( ) ( ) ( ) p y dy q x dx p y q q x p dz q dq p dp − − = − − = − − − − = = ⇒ . Taking first two members, we have Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 10 q dq p dp = Integrating on both sides, aq p a log q log p log = ⇒ + = . (ii) Putting p = aq in (i), we have 2 aq qy aqx = + a ax y q + = ⇒ . From (ii), we obtain ax y aq p + = = Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . ( ) ( ) dy a ax y dx ax y dz + + + = ∴ ( )( ) adx dy ax y adz + + = ⇒ Integrating on both sides, we get ( ) b ax y 2 1 az 2 + + = , which is the required complete solution involving two arbitrary constants a and b. General Integral: Writing ( ) a b φ = , we have ( ) ( ) a ax y 2 1 az 2 φ + + = (iii) Differentiating (iii) partially w.r.t. a, we have ( ) ( ) a ax y x z φ′ + + = . (iv) General integral is obtained by eliminating a from (iii) and (iv). Singular Integral: Differentiating the complete integral partially w.r.t. a and b, we have ( ) ax y x z + = and 0 = 1. Hence there is no singular integral. Q.No.9.: Solve the following non-linear partial differential equations by Charpit’s method: ( ) 0 q p qy px xy 2 2 2 = + + − − . Sol.: Given non-linear partial differential equation is ( ) 0 q p qy px xy 2 2 2 = + + − − . Here 0 xy 2 qy 2 px 2 q p f 2 2 = + − − + ≡ (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 11 q 2 y 2 dy p 2 x 2 dx x 2 q 2 dq y 2 p 2 dp − = − = + − = + − ⇒ q y dy p x dx x q dq y p dp − = − = + − = + − ⇒ q p y x dy dx q p y x dq dp − − + + = − − + + ⇒ dy dx dq dp + = + ⇒ ( ) ( ) a y q x p = − + − ∴ (ii) dy dx dq dp + = + , ( ) ( ) 0 y q x p = − + − Equation (i) can be written as ( ) ( ) ( ) 2 2 2 y x y q x p − = − + − (iii) Putting the values of ( ) y q − from (ii) in (iii), we have ( ) ( ) [ ] ( ) 2 2 2 y x x p a x p − = − − + − ( ) ( ) ( ) { } 0 y x a x p a 2 x p 2 2 2 2 = − − + − − − ⇒ ( ) { } [ ] 4 y x a 8 a 4 a 2 x p 2 2 2 − − − + = − , (Taking only +ve sign) ( ) { } ( ¸ ( ¸ − − + + = ⇒ 2 2 a y x 2 a 2 1 x p ∴ From (ii), ( ) { } ( ¸ ( ¸ − − + − = − 2 2 a y x 2 a 2 1 a y q ( ) { } ( ¸ ( ¸ − − − + = 2 2 a y x 2 a 2 1 y q . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . ( ) ( ) { }( ) dy dx a y x 2 2 1 dy dx 2 a ydy xdx dz 2 2 − − − + + + + = ∴ ( ) ( ) ( ) dy dx 2 a y x 2 1 dy dx 2 a ydy xdx 2 2 − ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ − − + + + + = Integrating on both sides, we have Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 12 ( ) ( ) ( ) ( ) b 2 a y x y x log 4 a 2 a y x 2 y x 2 1 y x 2 a 2 y 2 x z 2 2 2 2 2 2 2 + ( ( ¸ ( ¸ ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ | | ¹ | \ | − − + − − ¦ ) ¦ ` ¹ ¦ ¹ ¦ ´ ¦ − − − + + + + = which is the required complete solution involving two arbitrary constants a and b. Q.No.10.: Solve the following non-linear partial differential equations by Charpit’s method: 2 2 q p qy px z + + + = Sol.: Given non-linear partial differential equation is 0 q p qy px z f 2 2 = − − − − ≡ . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ ( ) ( ) ( ) ( ) q 2 y dy p 2 x dx q 2 y q p 2 x p dz q q dq p p dp − − = − − − = − − − − − − = + − − = + − ⇒ ( ) ( ) ( ) ( ) q 2 y dy p 2 x dx q 2 y q p 2 x p dz 0 dq 0 dp + − = + = + + + = − = ⇒ From first two members, we get 0 dp = and dq = 0. Integrating, we obtain p = a and q = b. Putting in (i), we get 2 2 b a by ax z + + + = , which is the required complete solution involving two arbitrary constants a and b. Q.No.11.: Solve the following non-linear partial differential equations by Charpit’s method: ( ) 1 q z p z 2 2 2 2 = + Sol.: Given non-linear partial differential equation is ( ) 1 q z p z 2 2 2 2 = + . Here 0 1 z q z p f 2 2 4 2 = − + ≡ . Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ ( ) ( ) ( ) ( ) 2 4 2 3 2 2 3 2 qz 2 q pz 2 p dz z q 2 z p 4 q dq z q 2 z p 4 . p 0 dp − − = + = + + ⇒ 2 4 qz 2 dy pz 2 dx − = − = . Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 13 Taking first two members, we have q dq p dp = . Integrating on both sides, we obtain a log q log p log + = aq p = ⇒ . Putting aq p = in ( ) 1 q z p z 2 2 2 2 = + , we get ( ) 1 z a z 1 q 2 2 2 2 + = ( ) 1 z a z 1 q 2 2 + = ⇒ ( ) 1 z a z a p aq p 2 2 + = ⇒ = ∴ . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . ( ) ( ) dy 1 z a z 1 dx 1 z a z a dz 2 2 2 2 + + + = ∴ ( ) dy adx dz 1 z a z 2 2 + = + ⇒ Integrating on both sides, we get ( ) b y ax 1 z a a 3 1 2 / 3 2 2 2 + + = + ( ) ( ) 2 4 3 2 2 b y ax a 9 1 z a + + = + ⇒ , Ans. which is the required complete solution involving two arbitrary constants a and b. Q.No.12.: Solve the following non-linear partial differential equations by Charpit’s method: 0 1 qy 2 px 2 q p 2 2 = + − − + . Sol.: Given non-linear partial differential equation is 0 1 qy 2 px 2 q p f 2 2 = + − − + ≡ . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ ( ) ( ) ( ) ( ) y 2 q 2 dy x 2 p 2 dx y 2 q 2 q x 2 p 2 p dz q 2 dq p 2 dp − − = − − = − − − − = − = − ⇒ Taking the first two members, we have q dq p dp = aq p a log q log p log = ⇒ + = ⇒ . Putting in (i), we get Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 14 0 1 qy 2 aqx 2 q q a 2 2 2 = + − − + ( ) ( ) 0 1 q y ax 2 q 1 a 2 2 = + + − + ⇒ ( ) ( ) ( ) { } ( ) 1 a 2 1 a 4 y ax 4 y ax 2 q 2 2 2 + + − + + + = ⇒ ( ) ( ) ( ) { } ( ) 1 a 1 a y ax y ax q 2 2 2 + + − + + + = ⇒ . (Taking the positive sign only) and ( ) ( ) ( ) { } 1 a 1 a y ax y ax a aq p 2 2 2 + ( ¸ ( ¸ + − + + + = = Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . ( ) ( ) ( ) { } ( ) ( ) dy adx 1 a 1 a y ax y ax dz 2 2 2 + + + − + + + = ∴ Putting t y ax = + , so that dt dy adx = + , we have, ( ) ( ) { } dt 1 a t t dz 1 a 2 2 2 ( ¸ ( ¸ + − + = + . Integrating on both sides, we get ( ) ( ) { } ( ) { } b 1 a t t log 2 1 a 1 a t 2 t 2 t z 1 a 2 2 2 2 2 2 2 + ( ¸ ( ¸ + − + + − + − + = + , where y ax t + = , which is the required complete solution involving two arbitrary constants a and b. Q.No.13.: Solve the following non-linear partial differential equations by Charpit’s method: ( ) 2 z qy p + = . Sol.: Given non-linear partial differential equation is ( ) 0 z qy p f 2 = + + − ≡ . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ ( ) ( ) ( )( ) ( ) ( ) ( ) z qy y 2 dy 1 dx y . z qy 2 . q 1 p dz z qy q 4 dq z qy p 2 dp + − = − − = + − − − = + = + ⇒ . Taking first and fifth members, we have 0 y dy p dp = + y a p a log y log p log = ⇒ = + ⇒ . Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 15 ∴From (i), we obtain ( ) 2 z qy y a + = z qy y a + = | | ¹ | \ | ⇒ y z y a q 2 / 3 − = ⇒ . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . dy y z y a dx y a dz 2 / 3 | | ¹ | \ | − + = ∴ dy y a adx zdy ydz + = + ⇒ . Integrating on both sides, we get ( ) b ay 2 ax yz + + = , which is the required complete solution involving two arbitrary constants a and b. Q.No.14.: Solve the following non-linear partial differential equations by Charpit’s method: ( ) 2 / 1 pq 1 z qy px + = + . Sol.: Given non-linear partial differential equation is ( ) 0 pq 1 z qy px f 2 / 1 = + − + ≡ . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ ( ) ( ) ........ pq 1 q q dq pq 1 p p dp 2 / 1 2 / 1 = + − = + − ⇒ Taking the first two members, we have q dq p dp = aq p a log q log p log = ⇒ + = ⇒ . Putting in (i), we have ( ) 2 / 1 2 aq 1 z qy aqx + = + ( ) ( ) 2 2 2 2 aq 1 z y ax q + = + ⇒ ( ) 2 2 2 2 az y ax z q − + = ( ) { } 2 2 az y ax z q − + = ⇒ ( ) { } 2 2 az y ax az p − + = ⇒ . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . ( ) ( ) { } 2 2 az y ax zdy azdx dz − + + = ∴ ( ) { } 2 2 az y ax dy adx z dz − + + = ⇒ Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 16 Putting u a y ax = + , we get ( ) { } 2 2 az au du a z dz − = ( ) 2 2 z u z 1 dz du − = ⇒ . Again put, vz u = , so that ( ) 2 2 2 z z v z 1 dz dv z v − = + ( ) 1 v dz dv z v 2 − = + ⇒ ( ) v 1 v dz dv z 2 − − = ⇒ ( ) v 1 v dv z dz 2 − − = ⇒ ( ) { }dv v 1 v z dz 2 + − − = ⇒ . Integrating on both sides, we get ( ) ( ) b 2 v 1 v v log 2 1 1 v 2 v z log 2 2 2 + − ( ¸ ( ¸ ) ` ¹ ¹ ´ ¦ − + − − − = ( ) ( ) { } b 1 v v log 2 1 1 v 2 v 2 v z log 2 2 2 = − + − − + + ⇒ , where a z y ax z u v + = = . which is the required complete solution involving two arbitrary constants a and b. Q.No.15.: Solve the following non-linear partial differential equations by Charpit’s method: ( ) ( ) 0 1 q p xy pq y x 2 2 2 2 = − − − − . Sol.: Here ( ) ( ) 0 1 q p xy pq y x f 2 2 2 2 = − − − − ≡ . (i) Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ ( ) ( ) ( ) pxy 2 q y x dx q p x ypq 2 dq q p y pqx 2 dp 2 2 2 2 2 2 + − − = − − − = − − ⇒ ( ) .......... qxy 2 p y x dy 2 2 = − − − = Using x, y, p, q as multipliers, we have Each fraction 0 qdy pdx ydq xdp + + + = ( ) ( ) 0 ydq qdy pdx xdp = + + + ∴ . Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 17 Integrating, we get a qy px = + x qy a p − = ⇒ . ∴ From (i), we have ( ) ( ) 0 1 q x qy a xy q x qy a y x 2 2 2 2 2 = − ( ( ¸ ( ¸ − − − | ¹ | \ | − − ( ) ( ) { } 0 1 xyq y qy a q y x x qy a 2 2 2 = − + − − − − ⇒ ( ) 0 1 xyq ay q x x qy a 2 2 = − + − − ⇒ ( )( ) 0 x yq x ay q x qy a 2 2 2 = − + − − ⇒ 0 x yq x q ay yq x y a q ax 2 2 2 2 2 2 2 = − + + − − ⇒ ( ) x y a y x qa 2 2 2 + = + ⇒ ( ) 2 2 2 y x a x y a q + + = ⇒ ( ) ( ) ( ) 2 2 2 2 2 2 y x a y x a y x a y x y a a x 1 p + − = ( ( ¸ ( ¸ + + − = ∴ . Since we know that qdy pdx dy y z dx x z dz + = ∂ ∂ + ∂ ∂ = . ( ) ( ) ( ) 2 2 2 2 y x a dy x y a dx y x a dz + + + − = ( ) ( ) 2 2 2 2 y x a ydx xdy y x ydy xdx a dz + − + + + = ⇒ Integrating on both sides, we have ( ) b x y tan a 1 y x log 2 a z 1 2 2 + + + = − , which is the required complete solution involving two arbitrary constants a and b. Q.No.16.: Solve the following non-linear partial differential equation by Charpit’s method: ( ) 0 qy q xy p yz = − + − or yz qy pq pxy = + + . Sol.: Here ( ) 0 qy q xy p yz = − + − . (i) Charpit’s auxiliary equations are Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 18 q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ . Here py x f = ∂ ∂ , q xy p f + = ∂ ∂ , z q px y f − + = ∂ ∂ , q p q f + = ∂ ∂ , y z f − = ∂ ∂ . ( ) ( ) ( ) [ ] ( ) ( ) q p dy q xy dx y p q q xy p dz y q z q px dq ) y ( p py dp + − = + − = + + + − = − + − + = − + ⇒ . From 1 st and 2 nd member, we obtain qy z q px dq 0 dp − − + = 0 dp = ⇒ a p = ⇒ . Now from (i), 0 yz qy pq pxy = − + + ( ) 0 yz q y a axy = − + + ⇒ ( ) y a ax z y y a axy yz q + − = + − = ⇒ ( ) y a ax z y q + − = ⇒ . Now consider ( ) dy y a ax z y adx qdy pdx dz + − + = + = ( ) dy y a a ax z y adx dz + − = − ⇒ (ii) Put t ax z = − , dtd adx dz = − ∴(ii) reduces to dy y a a 1 dy y a a y a dy y a yt dt | | ¹ | \ | + − = + − + = + = Integrating both sides, we get dy y a 1 a dy dt t 1 + − = ∫ ∫ ∫ + constant (= log c) c log ) y a log( a y t log + + − = ⇒ [ ] ax z t − = ( ) c log ) y a log( a y ax z log + + − = − ⇒ , is the required solution. This solution can also be written as ( )( ) y 2 be a y ax z = + − . Q.No.17.: Solve the following non-linear partial differential equation by Charpit’s method: 2 p xp q = + . Sol.: Here 2 p xp q = + . (i) Charpit’s auxiliary equations are Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 19 Let 2 p xp q ) q , p , z , y , x ( f − + = Here p x f = ∂ ∂ , p 2 x p f − = ∂ ∂ 0 y f = ∂ ∂ , 1 q f = ∂ ∂ , 0 z f = ∂ ∂ . Charpit’s auxiliary equations are q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = ∂ ∂ − ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ ( ) [ ] 1 dy p 2 x dx q p 2 x p dz 0 q 0 dq 0 p dp − = + − = + − − = × + = + ⇒ From 1 st and 5 th members dy p dp − = a log y p log + − = ⇒ y a p log − = ⇒ y e a p − = ⇒ y ae p − = ⇒ Also from (i), 2 p xp q = + y 2 2 y xe a axe q − − = + ⇒ y y 2 2 axe e a q − − − = ⇒ . Now consider ( )dy axe e a dx ae qdy pdx dz y y 2 2 y − − − − + = + = ( ) dy e a xdy dx e a dz y 2 2 y − − + − = Integrating on both sides, we obtain ( ) c dy e a x e d a z y 2 2 y + + = − − ∫ ∫ c e 2 a axe z y 2 2 y + − = ⇒ − − , is the required solution. Q.No.18.: Solve the following non-linear partial differential equation by Charpit’s method: ( ) ( ) 0 q z b 1 p p 2 = − + + . Sol.: Let 0 zq bq p p ) q , p , z , y , x ( f 3 = − + + = Here 0 x f = ∂ ∂ , 1 p 3 p f 2 + = ∂ ∂ , 0 y f = ∂ ∂ , z b q f − = ∂ ∂ , q z f − = ∂ ∂ . Consider Charpit’s Auxiliary equations q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = | | ¹ | \ | ∂ ∂ + ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ etc ..... q dq pq dp 2 = − = − ⇒ Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 20 From 1 st and 2 nd members, q dq p dp = . Integrating on both sides, we get c log q log p log + = cq p = ⇒ . Now from (i) 0 q ) z b ( p p 3 = − + + ( ) 0 q z b cq q c 3 3 = − + + ⇒ 0 z b c q c 2 3 = − + + ⇒ 3 2 c b c z q − − = ⇒ c . c b c z q − − = ⇒ . Now consider dy b c z . c 1 dx c 1 . b c z qdy pdx dz 2 / 3 − − + − − = + = ( ) dy c y c 1 dz b c z 2 / 3 2 / 1 + = − − ⇒ − . Integrating on both sides, we get ( ) a dy c 1 dx c 1 dz b c a 2 / 3 2 / 1 + + = − − ∫ ∫ ∫ − a c y c x b c z 2 2 / 3 + + = − − ⇒ , is the required solution. Q.No.19.: Solve the following non-linear partial differential equation by Charpit’s method: qz p 1 2 = + . Sol.: Let 0 qz p 1 ) q , p , z , y , x ( f 2 = − + = Here 0 x f = ∂ ∂ , p 2 p f = ∂ ∂ , 0 y f = ∂ ∂ , z q f − = ∂ ∂ , q z f − = ∂ ∂ . Consider Charpit’s Auxiliary equations q f dy p f dx q f q p f p dz z f q y f dq z f p x f dp ∂ ∂ − = ∂ ∂ − = | | ¹ | \ | ∂ ∂ + ∂ ∂ − = ∂ ∂ + ∂ ∂ = ∂ ∂ + ∂ ∂ etc ..... q dq pq dp 2 = − = − ⇒ From 1 st and 2 nd members, q dq p dp = . Integrating on both sides, we get c log q log p log + = cq p = ⇒ . Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 21 Now from (i) 0 qz 1 p 2 = − + 0 qz q c 2 2 = − ⇒ 2 2 2 c 2 c 4 z z q − ± = ⇒ . Now consider dy c 4 z z c 2 1 dx c 4 z z c 2 1 qdy pdx dz 2 2 2 2 2 | ¹ | \ | − ± + | ¹ | \ | − ± = + = dy c 2 1 dx c 2 1 c 4 z z dz 2 2 2 + = − ± ⇒ dy c 2 1 dx c 2 1 dz c 4 z z c 4 z z . c 4 z z 1 2 2 2 2 2 2 2 + = − − − ± ⇒ m m . Integrating on both sides, we get b dy c 2 1 dx c 2 1 c 4 dz c 4 z z 2 2 2 2 + + = | ¹ | \ | − ∫ ∫ ∫ m ( ) [ ] 2 2 B A ) B A ( B A − = − + ( ¸ ( ¸ | ¹ | \ | − + − − ± ⇒ 2 2 2 2 2 2 c 4 z z log c 2 c 4 z 2 z 2 z d y 2 cx 2 b y c 2 1 x c 2 1 c 4 2 2 + + = | ¹ | \ | + + = , where b c 4 d 2 = , ( ( ¸ ( ¸ − − − = − ∫ 2 2 2 2 2 2 2 a x log 2 a a x 2 x dx a x Q ( ¸ ( ¸ | ¹ | \ | − + − − ± ⇒ 2 2 2 2 2 2 c 4 z z log c 2 c 4 z 2 z 2 z d y 2 cx 2 + + = , is the required solution. *************************************************** **************************************** ******************************** Home Assignments Q.No.1.: Solve the following non-linear partial differential equation by Charpit’s method: 0 4 z 4 z q 9 z p 16 2 2 2 2 2 = − + + . Hint: Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 22 qz 8 z q 18 qz p 32 dq pz 8 z pq 18 z p 32 dp 3 2 2 3 + + = + + 2 2 2 2 2 2 z q 18 z p 32 dz qz 18 dy pz 32 dx + − = + − = − = 0 pdz 4 dy . 0 dx . 1 dq . 0 zdp 4 = + + + + , a pz 4 x = + , z 4 a x p − − = , ( ) 2 2 a x 4 1 z 1 z 3 2 q − − − = . Ans.: ( ) ( ) 1 z 4 9 b y 4 a x 2 2 2 = + − + − . Q.No.2.: Solve the following non-linear partial differential equation by Charpit’s method: ( ) ( ) 0 q z b q 1 p 2 = − + + . Hint: ( ) = + = − + + = = 1 q dx q z b p pq 3 dz q dq pq dp 2 2 2 pq 2 b z dy + + − , ) i ( pc q ) ii ( = , Sub 1 b cz q − − = . Ans.: ( ) [ ] a cy x 1 b z c 2 + + = − − ; a, c are arbitrary constants. Q.No.3.: Solve the following non-linear partial differential equation by Charpit’s method: 0 q px q 2 = − − . Hint: a q = , ( ¸ ( ¸ + ± = a 4 x x 2 1 p 2 m . Ans.: b ay a 4 x x log a 2 a 4 x 2 x 2 1 4 x z 2 2 2 + + ( ¸ ( ¸ ) ` ¹ ¹ ´ ¦ + + + + ± − = . Q.No.4.: Solve the following non-linear partial differential equation by Charpit’s method: 0 q yzp 2 = − . Ans.: b y a ax 2 z 2 2 2 + + = . Q.No.5.: Solve the following non-linear partial differential equation by Charpit’s method: ( ) 0 y x qx py pq 2 2 2 = + + + + . Ans.: ( ) ( ) { } 2 2 2 2 a y x y x 2 1 y ay x ax z 2 + − − + − + − = . Partial Differential Equations: Charpit’s Method Prepared by: Dr. Sunil, NIT Hamirpur (HP) 23 ( ) { } ( ) { } ( ¸ ( ¸ + + − + − + b a y x 2 y x 2 log 2 a 2 2 2 / 3 2 Q.No.6.: Solve the following non-linear partial differential equation by Charpit’s method: 2 p 3 q = . Ans.: b y x 3 ax z 2 + + = . Q.No.7.: Solve the following non-linear partial differential equation by Charpit’s method: pq z = . Ans.: b y a 1 ax z 2 + + = . Q.No.8.: Solve the following non-linear partial differential equation by Charpit’s method: q p zpq + = . Ans.: ( ) b a y x 1 a 2 z 2 + | ¹ | \ | + + = . *************************************************** **************************************** ********************************