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Page 2 At ?? =0, ?? 0 = -?? 2 4 ? -?? 2 4 = -?? 2 4 +?? 5 ??? 5 =0 ?? = -?? 2 4 ?? 2?? (?????? ) From (x), (xi) and (xii), characteristics are ?? =?? (2-?? ?? ) ?? =v2?? (1-?? ?? ) ?? = -?? 2 4 ?? 2?? To find the equation of integral surface, we eliminate ?? and ?? ?? from (x) and (xi) and put their values in (xii). and ?? ?? ?= v2(?? -v2?? ) v2?? -?? ?? ?= v2?? -?? v2 Using these values of ?? and ?? ?? , we get ?? = -?? 2 4 ?? 2?? =- (v2?? -?? ) 2 2 × (v2?? -2?? ) 2 (v2?? -?? ) 2 ???? =- 2 ?? (?? -v2?? ) 2 ???? =-(?? -v2?? ) 2 , which is the required surface. Edurev123 4. Linear P.D.E. of Second Order with Constant Coefficient 4.1 Solve : (?? ?? -?? ?? ' -?? ?? ?? )?? =(?? ?? ?? +???? -?? ?? )?????? ????? -?????? ????? where ?? and ?? ' represent ?? ?? ?? and ?? ?? ?? . (2009 : 15 Marks) Page 3 At ?? =0, ?? 0 = -?? 2 4 ? -?? 2 4 = -?? 2 4 +?? 5 ??? 5 =0 ?? = -?? 2 4 ?? 2?? (?????? ) From (x), (xi) and (xii), characteristics are ?? =?? (2-?? ?? ) ?? =v2?? (1-?? ?? ) ?? = -?? 2 4 ?? 2?? To find the equation of integral surface, we eliminate ?? and ?? ?? from (x) and (xi) and put their values in (xii). and ?? ?? ?= v2(?? -v2?? ) v2?? -?? ?? ?= v2?? -?? v2 Using these values of ?? and ?? ?? , we get ?? = -?? 2 4 ?? 2?? =- (v2?? -?? ) 2 2 × (v2?? -2?? ) 2 (v2?? -?? ) 2 ???? =- 2 ?? (?? -v2?? ) 2 ???? =-(?? -v2?? ) 2 , which is the required surface. Edurev123 4. Linear P.D.E. of Second Order with Constant Coefficient 4.1 Solve : (?? ?? -?? ?? ' -?? ?? ?? )?? =(?? ?? ?? +???? -?? ?? )?????? ????? -?????? ????? where ?? and ?? ' represent ?? ?? ?? and ?? ?? ?? . (2009 : 15 Marks) Solution: Clearly the Auxiliary equation is ?? 2 -?? -2 =0 ? ?? 2 -2?? +?? -2 =0 ? ?? (?? -2)+1(?? -2) =0 ? ?? =-1,2 ?? ?? =?? 1 (?? +2?? )+?? 2 (?? -?? ) where ?? ? 1 and ?? 2 are Arbitrary function. Particular Integral = 1 (?? +?? ' )(?? -2?? ' ) [(2?? 2 +???? -2?? 2 )sin????? -cos????? ] Putting ?? -?? =?? we get ?? =?? +?? ?= 1 (?? -2?? ' ) ??[2?? 2 +?? (?? +?? )-2(?? +?? ) 2 sin??? (?? +?? )-cos??? (?? +?? )]???? ?= 1 (?? -2?? ' ) ??[2?? 2 +(?? +?? )[?? -2?? -2?? |sin?(?? 2 +???? )-cos?(?? ' +???? )]???? ?= 1 (?? -2?? ' ) ??[2?? 2 +(?? +?? )(-?? -2?? )sin?(?? 2 +???? )-cos?(?? 2 +???? )]???? = 1 (?? -2?? ' ) ??[2?? +?? )(?? -?? )sin?(?? 2 +???? )-cos?(?? 2 +???? )]???? ?= 1 (?? -2?? ' ) ??-(2?? +?? )(?? -?? ) cos?(?? 2 +???? ) (2?? +?? ) +??cos?(?? 2 +???? )???? -??cos?(?? 2 +???? )???? = 1 (?? -2?? ' ) (?? -2?? )cos????? ](?? ' -2?? -2?? )cos??? (?? ' -2?? )???? -?? =?? +2?? ????(?? ' -4?? )cos?(?? ' ?? -2?? 2 )???? ?? [sin?(?? ' ?? -2?? 2 )] (?? ' -4?? ) (?? ' -4?? ) ?=sin?[(?? +2?? )?? -2?? 2 ] ?=sin????? Clearly, the required solution is ?? =?? 1 +?? 2 4.2 Solve the PDE: (?? ?? -?? ' )(?? -?? ?? ' )?? =?? ?? ?? +?? +???? (2010 : 12 Marks) Solution: Page 4 At ?? =0, ?? 0 = -?? 2 4 ? -?? 2 4 = -?? 2 4 +?? 5 ??? 5 =0 ?? = -?? 2 4 ?? 2?? (?????? ) From (x), (xi) and (xii), characteristics are ?? =?? (2-?? ?? ) ?? =v2?? (1-?? ?? ) ?? = -?? 2 4 ?? 2?? To find the equation of integral surface, we eliminate ?? and ?? ?? from (x) and (xi) and put their values in (xii). and ?? ?? ?= v2(?? -v2?? ) v2?? -?? ?? ?= v2?? -?? v2 Using these values of ?? and ?? ?? , we get ?? = -?? 2 4 ?? 2?? =- (v2?? -?? ) 2 2 × (v2?? -2?? ) 2 (v2?? -?? ) 2 ???? =- 2 ?? (?? -v2?? ) 2 ???? =-(?? -v2?? ) 2 , which is the required surface. Edurev123 4. Linear P.D.E. of Second Order with Constant Coefficient 4.1 Solve : (?? ?? -?? ?? ' -?? ?? ?? )?? =(?? ?? ?? +???? -?? ?? )?????? ????? -?????? ????? where ?? and ?? ' represent ?? ?? ?? and ?? ?? ?? . (2009 : 15 Marks) Solution: Clearly the Auxiliary equation is ?? 2 -?? -2 =0 ? ?? 2 -2?? +?? -2 =0 ? ?? (?? -2)+1(?? -2) =0 ? ?? =-1,2 ?? ?? =?? 1 (?? +2?? )+?? 2 (?? -?? ) where ?? ? 1 and ?? 2 are Arbitrary function. Particular Integral = 1 (?? +?? ' )(?? -2?? ' ) [(2?? 2 +???? -2?? 2 )sin????? -cos????? ] Putting ?? -?? =?? we get ?? =?? +?? ?= 1 (?? -2?? ' ) ??[2?? 2 +?? (?? +?? )-2(?? +?? ) 2 sin??? (?? +?? )-cos??? (?? +?? )]???? ?= 1 (?? -2?? ' ) ??[2?? 2 +(?? +?? )[?? -2?? -2?? |sin?(?? 2 +???? )-cos?(?? ' +???? )]???? ?= 1 (?? -2?? ' ) ??[2?? 2 +(?? +?? )(-?? -2?? )sin?(?? 2 +???? )-cos?(?? 2 +???? )]???? = 1 (?? -2?? ' ) ??[2?? +?? )(?? -?? )sin?(?? 2 +???? )-cos?(?? 2 +???? )]???? ?= 1 (?? -2?? ' ) ??-(2?? +?? )(?? -?? ) cos?(?? 2 +???? ) (2?? +?? ) +??cos?(?? 2 +???? )???? -??cos?(?? 2 +???? )???? = 1 (?? -2?? ' ) (?? -2?? )cos????? ](?? ' -2?? -2?? )cos??? (?? ' -2?? )???? -?? =?? +2?? ????(?? ' -4?? )cos?(?? ' ?? -2?? 2 )???? ?? [sin?(?? ' ?? -2?? 2 )] (?? ' -4?? ) (?? ' -4?? ) ?=sin?[(?? +2?? )?? -2?? 2 ] ?=sin????? Clearly, the required solution is ?? =?? 1 +?? 2 4.2 Solve the PDE: (?? ?? -?? ' )(?? -?? ?? ' )?? =?? ?? ?? +?? +???? (2010 : 12 Marks) Solution: Given, the equation is (?? 2 -?? ' )(?? -2?? ' )?? =?? 2?? +?? +???? Complementary Function : The auxiliary equation is : (?? 2 -1)(?? -2) =0 ? ?? =±1 ? Solution is ?? ?? =?? 1 Particular Integral : ?? ?? = 1 (?? 2 -?? ' )(?? -2?? ' ) ?? 2?? +?? + 1 (?? 2 -?? ' )(?? -2?? ' ) ???? ? ?? ?? = 1 (2 2 -1)(?? -2?? ' ) ?? 2?? +?? + 1 (?? 2 -?? ' )·?? × 1 (1- 2?? ' ?? ) ???? ? ?? ?? = 1 3 1 (?? -2?? ' ) ?? 2?? +?? + 1 (?? 2 -?? ' ) · 1 ?? (1+ 2?? ' ?? )???? ?? ?? ?? = 1 3 ?? 1·1! ?? 2?? +?? + 1 (?? 2 -?? ' ) · 1 ?? (???? +?? 2 ) ?? ?? ?? = ?? 3 ?? 2?? +?? + 1 (?? 2 -?? ' ) ×( ?? 2 ?? 2 + ?? 3 3 ) ? ?? ?? = ?? 3 ?? 2?? +?? + 1 ?? 2 ×(1- ?? ' ?? 2 ) -1 (?? 2 ?? + ?? 3 3 ) ? ?? ?? = ?? 3 ?? 2?? +?? + 1 ?? 2 ×(1+ ?? ' ?? 2 )(?? 2 ?? + ?? 3 3 ) ? ?? ?? = ?? 3 ?? 2?? +?? + 1 ?? 2 ×(?? 2 ?? + ?? 3 3 + ?? 4 12 ) ? ?? ?? = ?? 3 ?? 2?? +?? + 1 ?? ( ?? 3 ?? 3 + ?? 4 12 + ?? 5 60 ) ? ?? ?? = ?? 3 ?? 2?? +?? + ?? 4 ?? 12 + ?? 5 60 + ?? 6 360 Now, ?? =?? ?? +?? ?? ? ?? =?? 1 (?? +?? )+?? 2 (?? -?? )+?? 3 (?? +2?? )+ ?? 3 ?? 2?? +?? + ?? 4 ?? 12 + ?? 5 60 + ?? 6 360 4.3 Solve the PDE (?? ?? -?? ?? +?? +?? ?? ' -?? )?? =?? (?? -?? ) -?? ?? ?? (2011 : 12 Marks) Page 5 At ?? =0, ?? 0 = -?? 2 4 ? -?? 2 4 = -?? 2 4 +?? 5 ??? 5 =0 ?? = -?? 2 4 ?? 2?? (?????? ) From (x), (xi) and (xii), characteristics are ?? =?? (2-?? ?? ) ?? =v2?? (1-?? ?? ) ?? = -?? 2 4 ?? 2?? To find the equation of integral surface, we eliminate ?? and ?? ?? from (x) and (xi) and put their values in (xii). and ?? ?? ?= v2(?? -v2?? ) v2?? -?? ?? ?= v2?? -?? v2 Using these values of ?? and ?? ?? , we get ?? = -?? 2 4 ?? 2?? =- (v2?? -?? ) 2 2 × (v2?? -2?? ) 2 (v2?? -?? ) 2 ???? =- 2 ?? (?? -v2?? ) 2 ???? =-(?? -v2?? ) 2 , which is the required surface. Edurev123 4. Linear P.D.E. of Second Order with Constant Coefficient 4.1 Solve : (?? ?? -?? ?? ' -?? ?? ?? )?? =(?? ?? ?? +???? -?? ?? )?????? ????? -?????? ????? where ?? and ?? ' represent ?? ?? ?? and ?? ?? ?? . (2009 : 15 Marks) Solution: Clearly the Auxiliary equation is ?? 2 -?? -2 =0 ? ?? 2 -2?? +?? -2 =0 ? ?? (?? -2)+1(?? -2) =0 ? ?? =-1,2 ?? ?? =?? 1 (?? +2?? )+?? 2 (?? -?? ) where ?? ? 1 and ?? 2 are Arbitrary function. Particular Integral = 1 (?? +?? ' )(?? -2?? ' ) [(2?? 2 +???? -2?? 2 )sin????? -cos????? ] Putting ?? -?? =?? we get ?? =?? +?? ?= 1 (?? -2?? ' ) ??[2?? 2 +?? (?? +?? )-2(?? +?? ) 2 sin??? (?? +?? )-cos??? (?? +?? )]???? ?= 1 (?? -2?? ' ) ??[2?? 2 +(?? +?? )[?? -2?? -2?? |sin?(?? 2 +???? )-cos?(?? ' +???? )]???? ?= 1 (?? -2?? ' ) ??[2?? 2 +(?? +?? )(-?? -2?? )sin?(?? 2 +???? )-cos?(?? 2 +???? )]???? = 1 (?? -2?? ' ) ??[2?? +?? )(?? -?? )sin?(?? 2 +???? )-cos?(?? 2 +???? )]???? ?= 1 (?? -2?? ' ) ??-(2?? +?? )(?? -?? ) cos?(?? 2 +???? ) (2?? +?? ) +??cos?(?? 2 +???? )???? -??cos?(?? 2 +???? )???? = 1 (?? -2?? ' ) (?? -2?? )cos????? ](?? ' -2?? -2?? )cos??? (?? ' -2?? )???? -?? =?? +2?? ????(?? ' -4?? )cos?(?? ' ?? -2?? 2 )???? ?? [sin?(?? ' ?? -2?? 2 )] (?? ' -4?? ) (?? ' -4?? ) ?=sin?[(?? +2?? )?? -2?? 2 ] ?=sin????? Clearly, the required solution is ?? =?? 1 +?? 2 4.2 Solve the PDE: (?? ?? -?? ' )(?? -?? ?? ' )?? =?? ?? ?? +?? +???? (2010 : 12 Marks) Solution: Given, the equation is (?? 2 -?? ' )(?? -2?? ' )?? =?? 2?? +?? +???? Complementary Function : The auxiliary equation is : (?? 2 -1)(?? -2) =0 ? ?? =±1 ? Solution is ?? ?? =?? 1 Particular Integral : ?? ?? = 1 (?? 2 -?? ' )(?? -2?? ' ) ?? 2?? +?? + 1 (?? 2 -?? ' )(?? -2?? ' ) ???? ? ?? ?? = 1 (2 2 -1)(?? -2?? ' ) ?? 2?? +?? + 1 (?? 2 -?? ' )·?? × 1 (1- 2?? ' ?? ) ???? ? ?? ?? = 1 3 1 (?? -2?? ' ) ?? 2?? +?? + 1 (?? 2 -?? ' ) · 1 ?? (1+ 2?? ' ?? )???? ?? ?? ?? = 1 3 ?? 1·1! ?? 2?? +?? + 1 (?? 2 -?? ' ) · 1 ?? (???? +?? 2 ) ?? ?? ?? = ?? 3 ?? 2?? +?? + 1 (?? 2 -?? ' ) ×( ?? 2 ?? 2 + ?? 3 3 ) ? ?? ?? = ?? 3 ?? 2?? +?? + 1 ?? 2 ×(1- ?? ' ?? 2 ) -1 (?? 2 ?? + ?? 3 3 ) ? ?? ?? = ?? 3 ?? 2?? +?? + 1 ?? 2 ×(1+ ?? ' ?? 2 )(?? 2 ?? + ?? 3 3 ) ? ?? ?? = ?? 3 ?? 2?? +?? + 1 ?? 2 ×(?? 2 ?? + ?? 3 3 + ?? 4 12 ) ? ?? ?? = ?? 3 ?? 2?? +?? + 1 ?? ( ?? 3 ?? 3 + ?? 4 12 + ?? 5 60 ) ? ?? ?? = ?? 3 ?? 2?? +?? + ?? 4 ?? 12 + ?? 5 60 + ?? 6 360 Now, ?? =?? ?? +?? ?? ? ?? =?? 1 (?? +?? )+?? 2 (?? -?? )+?? 3 (?? +2?? )+ ?? 3 ?? 2?? +?? + ?? 4 ?? 12 + ?? 5 60 + ?? 6 360 4.3 Solve the PDE (?? ?? -?? ?? +?? +?? ?? ' -?? )?? =?? (?? -?? ) -?? ?? ?? (2011 : 12 Marks) Solution: The given partial differential equation is (?? 2 -?? 2 +?? +3?? -2)?? ?=?? ?? -?? -?? 2 ?? ? the compiementary function of (??) is ?? ?? =?? -2?? ?? 1 (?? +?? )+?? ?? ?? 2 (?? -?? ) (???? ) The particular integral of (i) is ?? ?? = 1 (?? -?? ' +2)(?? +?? ' -1) (?? ?? -?? -?? 2 ?? ) = 1 (?? -?? ' +2)(?? +?? ' -1) ?? ?? -?? - 1 (?? -?? ' +2)(?? +?? ' -1) ·?? 2 ?? = 1 (1+1+2)(1-1-1) ?? ?? -?? - 1 ?? 2 -?? 2 +?? +3?? ' -2 ·?? 2 ?? = -1 4 ?? ?? -?? + 1 2[1- ?? 2 -?? 2 +?? +3?? 2 ] ·?? 2 ?? = -1 4 ?? ?? -?? + 1 2 [1+( ?? 2 -?? 2 +?? +3?? ' 2 )+( ?? 2 -?? 2 +?? +3?? ' 2 ) 2 +?..]?? 2 ?? = -1 4 ?? ?? -?? + 1 2 [1+ ?? 2 + 3?? ' 2 +( 1 2 + 1 4 )?? 2 +( 9 4 - 1 2 )?? '2 + 3 2 ?? ?? ' + 21 8 ?? 2 ?? ' - 23 8 ?? ?? '2 …..]?? 2 ?? = -1 4 ?? ?? -?? + 1 2 [?? 2 ?? +???? + 3 2 ?? 2 + 3 2 ?? +3?? + 21 4 ] ??? =?? ?? +?? ?? =?? -2?? ?? 1 (?? +?? )+?? ?? ?? 2 (?? +?? )- 1 4 ?? ?? -?? + 1 2 [?? 2 ?? +???? + 3 2 ?? 2 + 3 2 ?? +3?? + 21 4 ] 4.4 Solve the partial differential equation: (?? -?? ?? ' )(?? -?? ' ) ?? ?? =?? ?? +?? (2011 : 12 Marks) Solution:Read More
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