Questions and Answers #4 Complex Variables

Questions and Answers Sheet 4 Complex Variables Question #1 The joint probability distribution of thr random variables X and Y is given below: cxy 0  x  2,0  y  x f ( x, y) =  other 0 a.Find the value of the constant. b.Calculate the covariance and the correlation of the X and Y random variables. c. Calculate the expected value of the random variable 𝑍 = 2𝑋 − 3𝑌 + 2 Answer: Step 1 Given: cxy 0  x  2,0  y  x f ( x, y) =  other 0 a. value of constant c, 2 𝑥 ∫0 ∫0 𝑐𝑥𝑦 𝑑𝑦 𝑑𝑥 = 1 𝑦2 2 𝑥 ∫0 𝑐𝑥 [ 2 ] 𝑑𝑥 = 1 𝑐 0 2 ∫ 𝑥 × 𝑥 2 𝑑𝑥 = 1 2 0 𝑐 𝑥4 2 2 [ ] =1 4 0 2c=1 1 𝑐= 2 Step 2 b. we have given a joint pdf, cxy 0  x  2,0  y  x f ( x, y) =  other 0 marginal pdf of x, 𝑥 𝑓1 (𝑥) = ∫0 𝑓(𝑥, 𝑦)𝑑𝑦 𝑥1 𝑓1 (𝑥) = ∫0 𝑥𝑦𝑑𝑦 2 𝑥 1 𝑦2 𝑓1 (𝑥) = 𝑥 × ( ) 2 2 0 1 3  x f1 ( x) =  4 0 0 x2 otherwise 2 𝐸(𝑥) = ∫ 𝑥 × 𝑓1 (𝑥)𝑑𝑥 𝐸(𝑥) = 𝐸(𝑥) = 𝐸(𝑥) = Step 3 0 2 ∫0 × 𝑥3 4 5 2 1 𝑥 ( ) 4 5 0 8 5 𝑑𝑥 2 𝐸(𝑥 2 ) = ∫0 𝑥 2 × 𝑓1 (𝑥)𝑑𝑥 𝑥3 2 𝐸(𝑥 2 ) = ∫0 𝑥 2 × 𝐸(𝑥 2 ) = ( ) 𝐸(𝑥 2) 4 8 = 6 𝑑𝑥 4 2 1 𝑥6 0 3 Variance of x is, 𝑉𝑎𝑟(𝑥) = 𝐸(𝑥 2 ) − [𝐸(𝑥)]2 8 8 2 3 200 5 𝑉𝑎𝑟(𝑥) = − [ ] 𝑉𝑎𝑟(𝑥) = 𝑉𝑎𝑟(𝑥) = − 75 8 192 75 75 Step 4 marginal pdf of y is, 𝑥 𝑓2 (𝑦) = ∫0 𝑓(𝑥, 𝑦)𝑑𝑥 2 1 𝑓2 (𝑦) = ∫𝑦 𝑥𝑦 𝑑𝑥 2 𝑦 𝑥2 2 2 𝑦 𝑦 𝑓2 (𝑦) = ( ) 2 𝑓2 (𝑦) = (4 − 𝑦 2 ) 4 y 2  (4 − y ) 0  y  2 f2 ( y) =  4 0 otherwise 2 𝐸(𝑦) = ∫0 𝑦 × 𝑓2 (𝑦)𝑑𝑦 2 𝑦 𝐸(𝑦) = ∫0 𝑦 × (4 − 𝑦 2 )𝑑𝑦 4 1 4𝑦 3 𝐸(𝑦) = [ 𝐸(𝑦) = 𝑦5 − 4 3 16 2 ] 5 0 15 Step 5 2 𝐸(𝑦 2 ) = ∫0 𝑦 2 × 𝑓2 (𝑦)𝑑𝑦 2 𝑦 𝐸(𝑦 2 ) = ∫0 𝑦 2 × × (4 − 𝑦 2 )𝑑𝑦 4 𝐸(𝑦 2) 1 4 = [𝑦 − 𝑦6 2 ] 4 1 6 0 64 4 1 32 6 𝐸(𝑦 2 ) = [16 − ] 𝐸(𝑦 2 ) = [ ] 𝐸(𝑦 2) = 4 4 6 3 Variance of x is, 𝑉𝑎𝑟(𝑦) = 𝐸(𝑦 2 ) − [𝐸(𝑦)]2 4 16 2 3 5 𝑉𝑎𝑟(𝑦) = − [ ] 𝑉𝑎𝑟(𝑥) = 0.1956 Step 6 Now, 𝐸(𝑥𝑦) = ∫ ∫ 𝑥𝑦 𝑓(𝑥, 𝑦)𝑑𝑦 𝑑𝑥 2 𝑥 1 𝐸(𝑥𝑦) = ∫0 ∫0 𝑥𝑦 𝑥𝑦 𝑑𝑦𝑑𝑥 2 1 2 1 2 𝑥 𝐸(𝑥𝑦) = ∫0 ∫0 𝑥 2 𝑦 2 𝑑𝑦 𝑑𝑥 2 𝑦3 𝑥 𝐸(𝑥𝑦) = ∫0 𝑥 2 ( ) 𝑑𝑥 2 3 0 2 1 𝐸(𝑥𝑦) = ∫0 𝑥 5 𝑑𝑥 6 2 1 𝑥6 𝐸(𝑥𝑦) = [ ] 𝐸(𝑥𝑦) = 𝐸(𝑥𝑦) = 6 6 0 1 36 16 [64] 9 Step 7 Covariance between x and y is, 𝐶𝑜𝑣(𝑥, 𝑦) = 𝐸(𝑥𝑦) − 𝐸(𝑥)𝐸(𝑦) 𝐶𝑜𝑣(𝑥, 𝑦) = 16 9 8 16 5 15 −( )( ) 𝐶𝑜𝑣(𝑥, 𝑦) = 0.0711 Correlation between x and y is, 𝐶𝑜𝑟𝑟(𝑥, 𝑦) = 𝐶𝑜𝑟𝑟(𝑥, 𝑦) = 𝐶𝑜𝑣(𝑥,𝑦) √𝑉(𝑥)×√𝑉(𝑦) 0.0711 √ 8 ×√0.1956 75 𝐶𝑜𝑟𝑟(𝑥, 𝑦) = 0.4924 c. Let 𝑧 = 2𝑥 − 3𝑦 + 2 𝐸(𝑧) = 𝐸(2𝑥 − 3𝑦 + 2) 𝐸(𝑧) = 2𝐸(𝑥) − 3𝐸(𝑦) + 2 8 16 5 15 𝐸(𝑧) = 2 ( ) − 3 ( ) + 2 𝐸(𝑧) = 2 Hence, the expected value of random variable z is 2. Question #2 Some of the solution sets for quadratic equations in the next sections in this chapter will contain complex numbers such as = = = −4+√−12 2 and −4−√−12 2 . We can simplify the first number as follows. −4+√−12 2 −4+𝑖√12 2 −4+2𝑖√3 2 2(−2+𝑖√3) 2 = −2 + 𝑖√3 Simplify each of the following complex numbers. 10+√−45 4 Answer: Step 1 Given: 10+√−45 4 Step 2 10+√−45 = = = 4 10+𝑖√9×5 10+𝑖√45 4 (√−1 = 𝑖) 4 10+𝑖3√5 4 Question #3 Some of the solution sets for quadratic equations in the next sections in this chapter will contain complex numbers such as −4+√−12 2 and −4−√−12 2 . We can simplify the first number as follows. −4+√−12 2 = = = −4+𝑖√12 2 −4+2𝑖√3 2 2(−2+𝑖√3) 2 = −2 + 𝑖√3 Simplify each of the following complex numbers. −1−√−18 2 Answer: Step 1 Some of the solution sets for quadratic equations in the next sections in this chapter will contain complex numbers such as −4+√−12 2 and −4−√−12 2 . We can simplify the first number as follows. −4 + √−12 −4 + 𝑖√12 = 2 2 −4+2𝑖√3 = = 2 2(−2+𝑖√3) 2 = −2 + 𝑖√3 Step 2 Simplify each of the following complex numbers. −1−√−18 = = = −1−𝑖√18 2 −1−𝑖3√2 2 2 −1 2 − 3√2 2 𝑖 Question #4 Solve the given differential equation by separation of variables. 𝑑𝑦 − (𝑦 − 8)2𝑑𝑥 = 0 Answer: Step 1 The differential equation is given as: 𝑑𝑦 − (𝑦 − 8)2𝑑𝑥 = 0 For finding the variable first we must put the same terms on one side and other terms on the other side of the equal to sign 𝑑𝑦 − (𝑦 − 8)2𝑑𝑥 = 0 𝑑𝑦 = (𝑦 − 8)2𝑑𝑥 𝑑𝑦 (𝑦−8) = 2𝑑𝑥 Step 2 Now integrating both sides we get, Where c is the integration constant variable. 𝑑𝑦 ∫ (𝑦−8) = ∫ 2𝑑𝑥 ln|(𝑦 − 8)| = 2𝑥 + 𝑐 Question #5 Find the Jacobian of the transformation 𝑥 = 𝑢 + 4𝑣, 𝑦 = 𝑢2 − 2𝑣 Answer: Basic information: given: x=u+4v 𝑦 = 𝑢2 − 2𝑣 x u Jacobian = y u 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢 = = 𝜕 𝜕𝑢 𝜕 𝜕𝑢 x v y v (𝑢 + 4𝑣) = 𝑢 (𝑢2 − 2𝑣) = 2𝑢 we thear v as command. 𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣 = = 𝜕 𝜕𝑣 𝜕 𝜕𝑣 (𝑢 + 4𝑣) = 4 (𝑢2 − 2𝑣) = −2 we thear u as command. Calculation of Jacobian Now, 𝜕𝑦 𝜕𝑢 𝜕𝑥 𝜕𝑢 = 2𝑢, = 1, 𝜕𝑦 𝜕𝑣 𝜕𝑥 𝜕𝑣 =4 = −2 Jacobian = 1 4 2u −2 = (−2)(1) − (4)(2𝑢) = −8𝑢 − 2 Jacobian = −8𝑢 − 2 Question #6 Evaluate the algebraic expressions for the given values of the variables. (𝑥 − 𝑦)2 , 𝑥 = 5 and 𝑦 = −3 Answer: Step 1 The given expression is (𝑥 − 𝑦)2 And, given values for which we have to find the value of the given expression are x=5 and y=-3 Step 2 So, to find the value of the given expression replace x with 5 and y with -3 in (𝑥 − 𝑦)2 2 (𝑥 − 𝑦)2 = (5 − (−3)) (𝑥 − 𝑦)2 = (5 + 3)2 (𝑥 − 𝑦)2 = 82 (𝑥 − 𝑦)2 = 8 × 8 (𝑥 − 𝑦)2 = 64 hence, (𝑥 − 𝑦)2 = 64 Question #7 Solve the system of equation by the method of your choice. 𝑥 2 + (𝑦 − 9)2 = 49 𝑥 2 − 7𝑦 = −14 Answer: Step 1 The system of equations is given by, 𝑥 2 + (𝑦 − 9)2 = 49(1) 𝑥 2 − 7𝑦 = −14(2) Step 2 Solve the system of equations. (7𝑦 − 14) + (𝑦 − 9)2 = 49 7𝑦 − 14 + 𝑦 2 − 18𝑦 + 81 − 49 = 0 𝑦 2 − 11𝑦 + 18 = 0 𝑦 2 − 9𝑦 − 2𝑦 + 18 = 0 𝑦(𝑦 − 9) − 2(𝑦 − 9) = 0 (𝑦 − 2)(𝑦 − 9) = 0 𝑦 = 2,9 Step 3 The solution of the system of equations is computed as follows. For y=2, the value of x is 𝑥 2 = 7(2) − 14 𝑥 2 = 14 − 14 𝑥2 = 0 x=0 For y=9, the value of x is 𝑥 2 = 7(9) − 14 𝑥 2 = 63 − 14 𝑥 2 = 49 𝑥 = ±7 The solution is {(0,2), (−7,9), (7,9)} Question #8 Convert the polar equation to rectangular coordinates. (Use variables x and y as needed. 𝑟= 3 1+sin(𝜃) Answer: Step 1 Consider the pola equation For converting polar coodinates (𝑟, 𝜃) into cartesian coordinates (x,y), use the following equations 𝑥 = 𝑟 cos(𝜃) 𝑦 = 𝑟 sin(𝜃) and 𝑥 2 + 𝑦 2 = 𝑟 2 3 Multiply both sides by 1 + sin(𝜃) into 𝑟 = 1+sin(𝜃) 𝑟(1 + sin(𝜃)) = 3 1+sin(𝜃) × (1 + sin(𝜃)) 𝑟(1 + sin(𝜃)) = 3 𝑟 + 𝑟 sin(𝜃) = 3 (Clear the paranthesis) Substitute 𝑟 = √𝑥 2 + 𝑦 2 and 𝑟 sin(𝜃) = 𝑦 𝑖𝑛𝑡𝑜 𝑟 + 𝑟 sin(𝜃) √𝑥 2 + 𝑦 2 + 𝑦 = 3 Step 2 Subtract y from both sides and further simplify it √𝑥 2 + 𝑦 2 + 𝑦 − 𝑦 = 3 − 𝑦 √𝑥 2 + 𝑦 2 = 3 − 𝑦 Squaring both sides of the equation √𝑥 2 + 𝑦 2 = 3 − 𝑦 2 (√𝑥 2 + 𝑦 2 ) = (3 − 𝑦)2 𝑥 2 + 𝑦 2 = 9 + 𝑦 2 − 6𝑦 (Use formula (𝑎 − 𝑏)2 = 𝑎2 + 𝑏 2 − 2𝑎𝑏 ) Subtract 𝑦 2 from both sides and further simplify it 𝑥 2 + 𝑦 2 − 𝑦 2 = 9 + 𝑦 2 − 𝑦 2 − 6𝑦 𝑥 2 = 9 − 6𝑦 Add 6y both sides and further simplify it 𝑥 2 + 6𝑦 = 9 − 6𝑦 + 6𝑦 𝑥 2 + 6𝑦 = 9 Which is required cartesian equation Question #9 solve the Initial Value Problem using Separation of Variables 𝜋 𝑦 ′ = cos 2 𝑥, 𝑦(0) = 4 Answer: Given differential equation, 𝑦 ′ = cos 2 𝑥 𝑑𝑦 = cos 2 𝑥 𝑑𝑥 𝑑𝑦 = cos 2 𝑥𝑑𝑥 Integrating , ∫ 𝑑𝑦 = ∫ cos 2 𝑥𝑑𝑥 Since ∫ cos 𝑥 𝑑𝑥 = sin 𝑥 sin 2𝑥 𝑦= 2 +𝐶 Given initial condition, 𝑦(0) = ∴ 𝜋 4 = 𝐶= sin 0 2 𝜋 𝜋 4 +𝐶 4 Hence solution is 𝑦= sin 2𝑥 2 + 𝜋 4 Question #10 Convert the equation to polar form. (Use variables r and 𝜃 as needed.) 𝑥 2 − 𝑦2 = 7 Answer: Step 1 We plug 𝑥 = 𝑟 cos 𝜃 𝑎𝑛𝑑 𝑦 = 𝑟𝑠𝑖𝑛 𝜃 𝑥 2 − 𝑦2 = 7 (𝑟 cos 𝜃)2 − (𝑟 sin 𝜃)2 = 7 𝑟 2 cos 2 𝜃 − 𝑟 2 sin2 𝜃 = 7 𝑟 2 (cos 2 𝜃 − sin2 𝜃) = 7 7 𝑟 2 = (cos2 2 𝜃−sin 𝜃) 7 2 𝑟 = cos(2𝜃) Step 2 Answer: 𝑟 2 = 7 cos(2𝜃) Question #11 Solve the given differential equation by separation of variables. csc(𝑦) 𝑑𝑥 + sec 2 (𝑥) 𝑑𝑦 = 0 Answer: Given: Given differential equation is csc(𝑦) 𝑑𝑥 + sec 2 (𝑥) 𝑑𝑦 = 0 We want to solve this differential equation by saperation of the variables Calculation: Given differential equation is csc(𝑦) 𝑑𝑥 + sec 2 (𝑥) 𝑑𝑦 = 0 ⇒ sec 2 (𝑥) 𝑑𝑦 = − csc(𝑦) 𝑑𝑥 𝑑𝑦 𝑑𝑥 ⇒ =− csc(𝑦) sec 2 (𝑥) This is variable saperable form so integrating both side 𝑑𝑦 𝑑𝑥 ∫ = −∫ csc(𝑦) sec 2 (𝑥) 1 1 Now csc(𝑦) = , sec(𝑥) = sin 𝑦 cos 𝑥 ⇒ ∫ sin(𝑦) 𝑑𝑦 = − ∫ cos 2 (𝑥) 𝑑𝑥 1 + cos(2𝑥) 𝑑𝑥 2 1 sin(2𝑥) ⇒ − cos(𝑦) = − (𝑥 + )+𝑐 2 2 1 sin(2𝑥) ⇒ cos(𝑦) = (𝑥 + ) + 𝑐 ′ ⋯ 𝑤ℎ𝑒𝑟𝑒 𝑐 ′ = −𝑐 2 2 This is required solution Answer: The solution of the differential equation csc(𝑦) 𝑑𝑥 + sec 2 (𝑥) 𝑑𝑦 = 0 ⇒ ∫ sin(𝑦) 𝑑𝑦 = − ∫ 1 sin(2𝑥) 2 2 is equal to cos(𝑦) = (𝑥 + + 𝑐′) Question #12 Write the augmented matrix for the system of linear equations. − x − 7 y + 8 z = 7  −8 x − 24 z = −35 7 x − y + 7 z = 15  Answer: Step 1 Given system of linear equation -x-7y+8z=7 -8x-24z=-35 7x-y+7z=15 We write augmented matrix Step 2 The above system can be written as -x-7y+8z=7 −8𝑥 + 0 ⋅ 𝑦 − 24𝑧 = −35 7x-y+7z=15 This system can be write Ax=B  −1 −7 8   x  7        Where A = −8 0 −24 , x = y , B = −35        7 −1 7   z   15   −1 −7 8  The augmented matrix =  A | B  = −8 0 −24   7 −1 7 7  −35 15  Question #13 Convert the polar equation to rectangular coordinates. (Use variables x and y as needed.) 𝑟 = 5(4 − sin(𝜃)) Answer: Step 1 We convert the Polar equation to rectangular coordinates 𝑟 = 5(4 − sin(𝜃)) Step 2 for polar equation to rectangular coordinates we put 𝑥 = 𝑟(cos(𝜃)), 𝑦 = 𝑟 sin(𝜃) Now 𝑥 2 + 𝑦 2 = 𝑟 2 (cos 2 (𝜃) + sin2 (𝜃)) ⇒ 𝑥 2 + 𝑦2 = 𝑟2 𝑦 and 𝜃 = tan−1 ( ) 𝑥 Now 𝑟 = 5(4 − sin(𝜃)) 𝑦 ⇒ 𝑟 = 5 (4 − ) 𝑟 ⇒ 𝑟 2 = 5(4𝑟 − 𝑦) ⇒ 𝑥 2 + 𝑦 2 = 5(4√𝑥 2 + 𝑦 2 − 𝑦) Hence the rectangular coordinates 𝑥 2 + 𝑦 2 = 5(4√𝑥 2 + 𝑦 2 − 𝑦) Question #14 Use the rules of exponents to simplify the expression. Write the answer with positive exponents. 2 2 (𝑥 3 ) 7 (𝑥 3 )3 Answer: Expression given is 2 2 (𝑥 3 ) 7 (𝑥 3 )3 2 = (𝑥)3 𝑥 ×2 7 3×3 4 = (𝑥)3 4 = 𝑥 3−7 𝑥7 4−21 3 =𝑥 =𝑥 −17 3 1 = 17 𝑥3 Question #15 (5−𝑗6)−(2−𝑗8) Given a complex number 𝑧 = (−3+𝑗4)(5−𝑗)+(4−𝑗6) Its magnitude is ____ and its phase is ____ (degress) Answer: Question 1 The objective is to find the magnitude and phase of given complex number. (5−𝑗6)−(2−𝑗8) 𝑧 = (−3+𝑗4)(5−𝑗)+(4−𝑗6) First try to simplify the given complex number into standard form of complex number, as (5−𝑗6)−(2−𝑗8) 𝑧 = (−3+𝑗4)(5−𝑗)+(4−𝑗6) 5−𝑗6−2+𝑗8 = (−3)(5−𝑗)+𝑗4(5−𝑗)+4−𝑗6 = = 3+𝑗2 −15+3𝑗+𝑗20+4+4−𝑗6 3+𝑗2 −7+𝑗17 (3+𝑗2)(−7−𝑗17) = (−7+𝑗17)(−7−𝑗17) = = = 3(−7−𝑗17)+𝑗2(−7−𝑗17) (−7)2 −(𝑗17)2 −21−𝑗51−𝑗14+34 49+289 −65𝑗+13 338 Now , magnitude of a complex number z=a+jb is |𝑧| = √𝑎2 + 𝑏 2 13 Here, 𝑎 = 338 ⇒ |𝑧| = √( = = 1 338 √26 26 , and 𝑏 = −65 2 13 338 338 ) +( −65 338 2 ) √(13 × 5)2 + (13)2 = = 13 338 √25 + 1 1 √26 And phase of a complex number z=a+jb is 𝜃 = tan−1 Here , 𝑎 = 13 338 ⇒ 𝜃 = tan−1 [ . and 𝑏 = −65 338 13 338 −65 𝑏 𝑎 338 ] = tan−1 (−5) ≈ −78.69 degree(approx) Thus , magnitude of z is 1 √26 and phase of z is -78.69 degrees. Question #16 Find the complex zeros of the following polynomial function. Write f in factored form. 𝑓(𝑥) = 𝑥 3 − 12𝑥 2 + 49𝑥 − 58 Answer: 𝑓(𝑥) = 𝑥 3 − 12𝑥 2 + 49𝑥 − 58 = 𝑥 3 − 2𝑥 2 − 10𝑥 2 + 20𝑥 + 29𝑥 − 58 = 𝑥 2 (𝑥 − 2) − 10𝑥(𝑥 − 2) + 29(𝑥 − 2) = (𝑥 − 2)(𝑥 2 − 10𝑥 + 29) 𝑓(𝑥) = (𝑥 − 2)(𝑥 2 − 10𝑥 + 29) for roots , f(x)=0 (𝑥 − 2)(𝑥 2 − 10𝑥 + 29) = 0 x=2 𝑥 2 − 10𝑥 + 29 = 0 a=1 , b=-10 ,c=29 𝑥= = 10±√100−116 2⋅1 10±√100−116 2 = 10±√−16 2 10 ± 4𝑖 = = 5 ± 2𝑖 2 x=5+2i ; 5-2i All the roots are x=2 , x=5+2i , x=5-2i complex roots are 5+2i , 5-2i Question #17 Write the system of linear equations represented by the augmented matrix. (Use variables x, y, z, and w, if applicable.) 1 −1 3 0 1 −1  0 0 1 0 3  −3 Use back-substitution to solve the system Answer: 1 −1 3  Given Augmented matrix is 0 1 −1  0 0 1 Let x ,y and z be the three variable Hence, solve AX=B 1 −1 3   x   0  0 1 −1  y  =  3       0 0 1   z   −3 Hence , sustem of equation is x-y+3z=0 y-z=3 z=-3 Now by Substitution back to find x,y and z Substitute z=-3 into y-z=3 y-(-3)=3 y+3=3 y=0 Substitute y=0 and z=-3 into x-y+3z=0 x-0+3(-3)=0 x-9=0 x=9 Therefore, (x,y,z)=(9,0,-3) Question #18 0 3  −3 Write the system of linear equations represented by the augmented matrix. (Use variables x, y, z, and w, if applicable. 1 1  4 −8  8 −1 Answer: Step 1 Given augmented matrix 1 1 [ AB] =  4 −8 8 −1 Step 2 ∴ Corresponding system of equation, x+y=8 (1) 4x-8y=-1 (2) From (1, y=8-x Substituting in (2) 4x-8(8-x)=-1 4x-64+8x=-1 12x=64-1 63 12𝑥 = 63 ⇒ 𝑥 = 12 Substituting in (1), 𝑦 = 8 − 𝑦= 33 63 12 12 Question #19 Solve the differential equation..(b and a are variables) (𝐷2 + 4𝐷 + 3)𝑏 = 1 + 2𝑎 + 3𝑎2 Answer: Given D.E is (𝐷2 + 4𝐷 + 3)𝑏 = 1 + 2𝑎 + 3𝑎2 (1) ⇒ 𝑏 ′′ + 4𝑏 ′ + 3𝑏 = 1 + 2𝑎 + 3𝑎2 The auxilary equation is 𝑚3 + 4𝑚 + 3 = 0 𝑚2 + 3𝑚 + 𝑚 + 3 = 0 𝑚(𝑚 + 3) + 1(𝑚 + 3) = 0 (𝑚 + 3)(𝑚 + 1) = 0 m=-3,-1 𝑏𝑐 = 𝑐1 𝑒 −3𝑎 + 𝑐2 𝑒 −𝑎 is a complimentary solution of (1) Now For 𝑏𝑝 = 𝐴𝑎2 + 𝐵𝑎 + 𝑐 be the particular solution by using method of undelemined coefficient 𝑏𝑝′ = 2𝐴𝑎 + 𝐵 𝑏𝑝′′ = 2𝐴 Then (1) becomes as 𝑏𝑝′′ + 4𝑏𝑝′ + 3𝑏𝑝 = 1 + 2𝑎 + 3𝑎2 2𝐴 + 4(2𝐴𝑎 + 𝐵) + 3(𝐴𝑎2 + 𝐵𝑎 + 𝐶) = 1 + 2𝑎 + 3𝑎2 3𝐴𝑎2 + (8𝐴 + 3𝐵)𝑎 + (2𝐴 + 4𝐵 + 3𝐶) = 1 + 2𝑎 + 3𝑎2 Comparing both sides 3𝑎 = 3 ⇒ 𝐴 = 1 8𝐴 + 3𝐵 = 2 ⇒ 8(1) + 3𝐵 = 2 3𝐵 = 2 − 8 −6 = −2 3 B=-2 2A+4B+3C=1 𝐵= 2(1) + 4(−2) + 3𝐶 = 1 ⇒ 𝐶 = Then 𝑏𝑝 = 𝑎2 − 2𝑎 + 7 7 3 3 The general solution is 𝑏 = 𝑏𝑐 + 𝑏𝑝 𝑏 = 𝑐1 𝑒 −3𝑎 + 𝑐2 𝑒 −𝑎 + 𝑎2 − 2𝑎 + 7 3 Is required solution Question #20 State wether the given equation is true for all values of the variables. (𝑥 + 𝑦)3 = 𝑥 3 + 𝑦 3 Answer: Step 1 We know, (𝑥 + 𝑦)3 = 𝑥 3 + 𝑦 3 + 3𝑥𝑦(𝑥 + 𝑦) = 𝑥 3 + 𝑦 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 = 𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 ⇒ (𝑥 + 𝑦)3 = 𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3 Given statement is: (𝑥 + 𝑦)3 = 𝑥 3 + 𝑦 3 Which is false as: (𝑥 + 𝑦)3 = 𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦62 + 𝑦 3 Hence , not true for all values of the variables. Step 2 Result: Given stetment is false Question #21 State wether the given equation is true for all values of the variables. (Disregard any value that make a denominator 0.) 1 + √𝑎 1 = 1−𝑎 1 − √𝑎 Answer: Step 1 Two equation is said to be true or correct if the left hand side and the right hand side of the equation are the same. Step 2 Now consider the equation given by; 1+√𝑎 1−𝑎 = 1 1−√𝑎 rewrite the denominator of the left hand side of the equation using difference of square formula as follows; 1+√𝑎 1−𝑎 = 1 1−√𝑎 1+𝑠𝑞𝑟𝑡𝑎 2 1−(√𝑎) = 1 1−√𝑎 1+√𝑎 (1+√𝑎)(1−√𝑎) 1 1−√𝑎 1 1−√𝑎 ⋅ 1+√𝑎 (1+√𝑎) 1 = = = 1 1−√𝑎 ⋯ [𝑎2 − 𝑏 2 = (𝑎 − 𝑏)(𝑎 + 𝑏)] 1 1−√𝑎 1−√𝑎 The above results stands for all real values of a except those values for which the denominator becomes zero. Hence proved.