Chapter 6: Functions with severable variables and Partial Derivatives:

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Chapter 6: Functions with severable variables and Partial Derivatives: Functions o several variables: A unction involving more than one variable is called unction with severable variables. Eamples: y (, y = ; +y (, y = ; y + y (, y = ; y (, y = + y. (, y, z = y ln( + z ; (, y, z = z + + y ; (, y, z = ye yz ; (, y, z = y + z + cos( yz.

Partial Derivatives: (,y ( + h, y (, y = = lim h 0 h Eample : Sol.: (, y + h (, y y = = lim y h 0 h (, y = + y y =?, =? y = 6 + y Eample : = y y y (, y = e sin( + y =?, =? y Sol.: y y = ye sin( + y + e cos( + y = e y sin( + y + e y cos( + y y

Partial Derivatives (,y,z: (, y, z = y = y Eample : Sol.: z = z (, y, z = ln( + y + z (, y, z = +y +z y y (, y, z = + y + z z z (, y, z = + y + z =?, =?, =? y z

Higher order Partial Derivatives: (, y Higher order Partial Derivatives: (, y, z?

(, y, z ((, y = =! = ( = = y! y y ( = = z! z z ( y = = y! y = y ( y = = yy! y y y z ( y = zy = yz! ( z = = z! z = z ( z = = zy! z yz y ( z = = zz! z z

Eample : Find all the second partial derivatives o Sol.: y (, y = e sin( + y y = [ y sin( + y + cos( + y ]e y = [ sin( + y + cos( + y]e y y y = [ y cos( + y sin( + y]e + y[ y sin( + y + cos( + y]e y = [ y sin( + y + ( + y cos( + y]e = y y y y = [ cos( + y sin( + y]e + [ sin( + y + cos( + y]e y

Eample 4:, yy, zz Find the second partial derivatives o the ollowing unction (, y, z = ln( + y + z Sol.: y y (, y, z = + y + z (, y, z = + y + z (, y, z = yy (, y, z = zz (, y, z = ( + y + z ( ( +y +z ( + y + z y( y ( +y +z ( + y + z z ( z ( +y +z z z (, y, z = + y + z = = = y + z ( +y +z + z y ( +y +z + y z ( +y +z

Eample 5: Find all the second partial derivatives o the ollowing unction (, y = y y + y + Sol.: (, y = y 4 y + y (, y = 6 y 4 y y (, y = 4 y + y yy (, y = 4 + 6 y y (, y = 8y + y = y (, y

Eample 6: Find the partial derivatives wy, wy, wy, w at (,0 o the unction y w = ye. Sol.: w = ye wy = [ e y y y y - y e y wy = e + ye ( ] [ ye y y y - ye. y + y e ( ]. y = e ( y ye ( y. y = e [( y y ( y ] y y = e [ y y + y ] = e [ 4 y + y ] wy = [ e y y y + e ( y] [ ye y + ye y ( y]. y = e ( y ye ( y. y y = e [( y y ( y ] = e [ 4 y + y ] w (,0 = 0 wy (,0 = 4 0 = 4 wy (,0 = ( = 4 = wy (,0

Chain Rule: Given a unction w=(,y,z, =g(s,t, y=h(s,t, z=k(s,t Eample : s w w w = + y + z, =, y = s + ln t, z = s Find and? t s t Sol.:

Eample : w w Find and in the ollowing cases : s t w = sin y, = s + t, y = st. w = + y, = s ln t, y = s + t. w = ln y, = s + t, y = st. 4 w = cos y, = s t, y = s. 5 w = y + yz, = s t, y = s t, z = s + t. Solution: w w w y w w y =. +. = sin y, = s, = cos y, = t. s s y s s y s w Thus, = sin y.( s + cos y.(t = s sin( st + t ( s + t cos( st. s w w w y y =. +. = t, = s. t t y t t t w Thus, = sin y.( t + cos y.( s = t sin( st + s( s + t cos( st. t

w w w y w w y =. +. = + y, = ln t, =, =. s s y s s y s w Thus, = ( + y.(ln t +.( s = (s ln t + 4s + t ln t + 4s ln t = [ s ln t + 4s + t ] ln t. w w w y =. +. t t y t s y =, =. t t t w s Thus, = ( + y.( +.( t t s = (s ln t + 4s + t + s ln t t,4,5 Quizes.

Dierentiation o Implicit unctions: Assume that z is an implicit unction o and y, that is, F(,y,z=0. Then F Fy z y = =. F y Fz z F F z = =. F Fz z z z Eample : I z y + yz + y = 0. Find and? y Solution: Let F (, y, z = z y + yz + y. Then F = y, Fy = + z + y, Fz = z + y. z y y z + z + y z y = =, = =. z + y z + y y z + y z + y

Eample : z z Find and in the ollowing cases : y sin y + z = yz. z yz + y + 4 z = 0. z + y 4 y z + y =. yz z y 4 e ye + ze =. 5 y + z + cos( yz = 4. Solution: Let F (, y, z = sin y + z yz. Then F = sin y yz, Fy = cos y z, Fz = z y. Fy F z sin y yz z cos y z = =, = =. Fz z y y Fz z y,,4,5 Quizes.

Chapter 7: Introductory to Dierential Equations: Deinition: A dierential equation (D.E. is an equation involving variables and derivatives. Eamples: y ' = 4 ( y' ' ' y = e y ( 4 5 = y The variable y is called the solution o the dierential equation. For eample the unction y = + C is a solution o the irst one.

Separable Dierential Equations: Deinition: A dierential equation (D.E. is called separable i it can be written in the orm: M(d+N(ydy=0. Eample : Sol.: Solve the dierential equation dy y e + =0 d y e d + dy = 0 e d + dy = 0, y e =C y y = e C dy y e + =0 d ( y 0

Eample : Sol.: Solve the dierential equation dy + y d = 0 dy + y d = 0, Divide by and y, 0, y 0. dy + d = 0 y y dy + d = 0 y + =C = C y = =y= y C C Remark: This solution is called the general solution o the dierential equation. I we add to the previous eample the boundary condition y(=-. Then substitute the values o and y into the general solution y= C we get C = and so the obtained solution is called the particular solution. y=

Linear First order Dierential Equations: Deinition: A dierential equation (D.E. is called Linear First order Dierential Equations i it can be written in the orm: y '+ P( y = Q( Eample : The ollowing dierential equations are L.F.O.D.E.:

Solving a First Order Linear Dierential Equation dy Put the equation into standard orm : + P ( y = Q (. d Identiy P( and Q(. Find P( d. P ( d 4 Let ρ ( = e. y= Q ( ρ ( d + C ρ (

Eample : Solve the dierential equation: y '+ y = 0 Solution: y '+ y = 0, divide by 0 y '+ y =, y '+ P( y = Q(. ρ ( = e P ( d with =e d P( =, Q( =, =e ln =. e n ln n =. 5 ρ ( Q( d = ( d = d = 5 + C. 4 Thus 5 C y= ρ ( Q( d + C = + C = +. ρ ( 5 5 [ ]

Eample : Solve the ollowing dierential equations: y '+ y = e. π y '+ y = sin, y ( =. y ' y = ( > 0. dy 4 y =, y (0 =. d dy 5 + y = 0. d + e Solution: ρ ( = e 6 y '+ y =, y (0 =. 7 y '+ y = cos(e. dy 8 + y =, y ( =. d dy 9 + 0 y =. d y '+ P( y = Q(. d P( =, Q( = e, =e. ρ ( Q( d = e y= ρ ( with (e d = e d = e + C. ρ ( Q ( d + C = e + C = e + Ce. e [,,4,5,6,7,8,9 Quizes. ] (

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