**Formation of partial differential equations:**

**There are two methods to form a partial differential equation.**

**By elimination of arbitrary constants.**

**By elimination of arbitrary functions.**

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**Formation of partial differential equations by elimination of arbitrary constants:**

**Let f(x, y, z, a, b) = 0 be an equation which contains two arbitrary constants ‘a’ and ‘b’.**

**Therefore partially differentiate the equation w.r.t. x and y to eliminate the constant ‘a’ and ‘b’**

** **

**Note: we use following notations in PDE**

**p = ∂z/∂x q = ∂z/∂y r = ∂ ^{2}z/∂x^{2 }s = ∂^{2}z/∂x∂y t = ∂^{2}z/∂y^{2 }**

**Types of solution:**

**(a) A solution in which the number of arbitrary ^{ }constants is equal to number of independent variables is called complete integral or complete solution.**

**(b) In complete integral if we give particular values to the arbitrary constants we get particular integral.**

**(c) Singular integral: let f(x, y, z, p, q) = 0 be a partial differential equation whose complete integral is Φ(x, y, z, a, b) = 0 …………….. (1)**

**Differentiating (1) partially w.r.t a and b and then equate to zero, we get**

** ∂Φ/∂a = 0……………. (2)**

** ∂Φ/∂b = 0 ……………. (3)**

**Eliminate a and b by using (1), (2) and (3).**

**The eliminate of a and b is called singular integral.**

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**Types of first order non-linear partial differential equations:**

**Type 1: F (p, q) = 0**

**Type 2: z = px +qy + f (p, q) [Caliraut’s form]**

**Type 3: F (z, p, q) = 0**

**Type 4: F _{1} (x, p) = F_{2}(y, q)**

**Type 5: F (x ^{m}p, y^{n}q) = 0 and F (z, x^{m}p, y^{n}q) = 0**

**Type 6: f (z ^{m}p, z^{m}q) = 0 and F_{1} (x, z^{m}p) = F_{2}(y, z^{m}q)**

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**Lagrange’s linear equations:**

**The equation of the form **

** Pp + Qq = R …………. (1)**

**Is known as Lagrange’s equation, where P, Q and R are functions of x, y and z. To solve this equation it is enough to solve the subsidiary equations.**

** dx/P = dy/Q = dz/R ……………(2)**

**If the solution of the subsidiary equation is of the form u(x, y) = c _{1} and v(x, y) = c_{2} then the solution of the given Lagrange’s equation is Φ(u, v) = 0.**

**To solve the subsidiary equations we have two methods:**

**1 M****ethod of Grouping:** ** Consider the subsidiary equation dx/P = dy/Q = dz/R. take any two members say first two or last two or first and last members. Now consider the first two members dx/P = dy/Q. If P and Q contain z(other than x and y) try to eliminate it. Now direct integration gives u(x, y) = c _{1. }Similarly take another_{ }two members dy/Q = dz/R. If Q and R contain x(other than y and z) try to eliminate it. Now direct integration gives v(y, z) = c_{2 . }Therefore solution of the given_{ }Lagrange’s equation is Φ(u, v) = 0.**

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** ****Method of multiplier’s** ** Choose any three multipliers l, m, n may be constants or function of x, y and z such that in**

** dx/P = dy/Q = dz/R = (ldx + mdy + ndz) / (lP + mQ + nR)**

** the expression lP + mQ + nR = 0**

**Hence ldx + mdy + ndz = 0 **

**[ since each of the above ratios equal to a constant**

** dx/P = dy/Q = dz/R = (ldx + mdy + ndz) / (lP + mQ + nR) = k (say)**

** (ldx + mdy + ndz) = k (lP + mQ + nR)**

**If lP + mQ + nR = 0 then ldx + mdy + ndz = 0]**

**Now direct integration gives u(x, y, z) = c _{1.}**

**similarly choose another set of multipliers l′, m′, n′**

**dx/P = dy/Q = dz/R = (l′dx +m′dy + n′dz) / (l′P + m′Q + n′R)**

** the expression l′P + m′Q + n′R = 0**

**therefore l′dx +m′dy + n′dz = 0 (as explained earlier)**

**Now direct integration gives v(x, y, z) = c _{2.}**

**Therefore solution of the given _{ }Lagrange’s equation is Φ(u, v) = 0.**

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**Homogeneous Linear partial differential equations:**

**F (x, y) = [a _{0} D^{n} + a_{1} D^{n-1} D′ + a_{2} D^{n-2 }D′^{2}_{ }+ ……. + a_{n}D′^{n}]z where D = ∂/∂x and D′ = ∂/∂y**

**Solution of Homogeneous Linear partial differential equations:**

**1 ****The Complementary function.** ** **

**The particular integral.**

**To find the Complementary function (C.F.):**

**The complementary function is the solution of the equation**

** a _{0} D^{n} + a_{1} D^{n-1} D′ + a_{2} D^{n-2 }D′^{2}_{ }+ ……. + a_{n}D′^{n }= 0.**

**In this equation , put D = m and D′ = 1 then we get an equation, which is called auxiliary equation**

**Hence the auxiliary equation is **

**a _{0} m^{n} + a_{1} m^{n-1} + a_{2} m^{n-2 }+ ……. + a_{n }^{ }= 0.**

**Let the root of this equation be m _{1, }m_{2, }m_{3,………..} m_{n.}**

**Case 1: If the roots are real or imaginary and different say m _{1} ≠ m_{2} ≠ m_{3} ≠…….. ≠ m_{n}. then the C.F. is**

** Z = f _{1} (y + m_{1}x) + f_{2} (y + m_{2}x) +……… + f_{n} (y + m_{n}x) **

**Case 2: If any two roots are equal, say m _{1} = m_{2} = m, and others are different then the C.F. is**

** Z = f _{1} (y + mx) + xf_{2} (y + mx) + f_{3} (y + m_{3}x) +……… + f_{n} (y + m_{n}x) **

** **** ****Case 3: If three roots are equal, say m _{1} = m_{2} = m_{3 }= m, then the C.F. is**

** Z = f _{1} (y + mx) + xf_{2} (y + mx) + x^{2}f_{3} (y + mx) +……… + f_{n} (y + m_{n}x) .**

** **

** ****To find the Particular Integral:**

**Rule1: If F(x, y) = e ^{ax+by }then**

** P.I. = 1/ Φ (D, D′). e ^{ax+by }**

** = 1/ Φ (a, b). e ^{ax+by }provided^{ }Φ (a, b) ≠ 0 [Replace D by a and D′ by b]**

** If Φ (a, b) = 0 refer rule 4.**

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**Rule2: If F(x, y) = sin (mx + ny) or cos (mx + ny) then **

** P.I. = 1/ Φ (D, D′). sin (mx + ny) or cos (mx + ny)**

** Replace D ^{2} by –m^{2}, D′^{2} by –n^{2} and DD′ by –mn in provided the denominator **

**is not equal to zero. If the denominator is zero refer rule 4.**

** **** ****Rule3: If F(x, y) = x ^{m} y^{n}**

** P.I. = 1/ Φ (D, D′). x ^{m} y^{n}**

^{ }= [Φ (D, D′)]^{-1 }x^{m} y^{n}

**Expand [Φ (D, D′)] ^{-1 }by using binomial theorem and then operate on x^{m} y^{n}**

**Note: 1/D denotes integration ^{ }w.r.t x, 1/ D′ denotes integration w.r.t y.**

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**Rule4: If F(x, y) is any other function, resolve Φ (D, D′) in to linear factor say (D – m _{1} D′) (D – m_{2} D′) etc. then the **

**P.I. = 1/(D – m _{1} D′) (D – m_{2} D′) F(x, y)**

** Note: 1/(D – mD′) F(x, y) = ∫ F(x, c-mx) dx, where y = c-mx.**

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**Note: If the denominator is zero in rule (1) and (2) then apply Rule (4)**