Equations #5 - Questions and Answers

Question 1 Find the given equation's general solution. 𝑦 ′ ′ − 4 𝑦 = 0 y ′′ −4y=0 Answer Step 1 To determine: The general solution of the given differential equation: 𝑦 ′ ′ − 4 𝑦 = 0 y ′′ −4y=0 Step 2 Concept used: Consider a differential equation: 𝑦 ′ ′ + 𝑎 𝑦 ′ + 𝑏 𝑦 = 0 y ′′ +ay ′ +by=0 Write the characteristic (or auxiliary) equation: 𝑟 2 + 𝑎 𝑟 + 𝑏 = 0 r 2 +ar+b=0 Solve this and find the roots 𝑟 1 r 1 ​ and 𝑟 2 r 2 ​ . General solution of the equation: 𝑦 ( 𝑥 ) = 𝐶 1 𝑒 𝑟 1 𝑥 + 𝐶 2 𝑒 𝑟 2 𝑥 y(x)=C 1 ​ e r 1 ​ x +C 2 ​ e r 2 ​ x Where 𝐶 1 C 1 ​ and 𝐶 2 C 2 ​ are the constants. Step 3 Explanation: Given that: 𝑦 ′ ′ − 4 𝑦 = 0 y ′′ −4y=0 Write the characteristic (or auxiliary) equation: 𝑟 2 − 4 = 0 r 2 −4=0 Solve it: ( 𝑟 + 2 ) ( 𝑟 − 2 ) = 0 (r+2)(r−2)=0 𝑟 = ± 2 r=±2 The general solution of the equation is: 𝑦 ( 𝑥 ) = 𝐶 1 𝑒 2 𝑥 + 𝐶 2 𝑒 − 2 𝑥 y(x)=C 1 ​ e 2x +C 2 ​ e −2x Question 2 Solve each equation for 𝑥 x: 3 𝑎 𝑡 𝑥 − 𝑐 𝑑 𝑥 = 2 𝑎 𝑏 − 2 𝑐 3atx−cdx=2ab−2c Answer Use the Distributive Property: 𝑎 ( 3 𝑡 𝑥 − 𝑐 𝑑 ) = 3 𝑎 𝑡 𝑥 − 2 𝑎 𝑏 a(3tx−cd)=3atx−2ab and  𝑐 ( 𝑑 𝑥 − 2 ) = 𝑐 𝑑 𝑥 − 2 𝑐 and c(dx−2)=cdx−2c So, the equation becomes: 3 𝑎 𝑡 𝑥 − 2 𝑎 𝑏 = 𝑐 𝑑 𝑥 − 2 𝑐 3atx−2ab=cdx−2c We want to collect all terms with 𝑥 x on the left side, so subtract 𝑐 𝑑 𝑥 cdx from both sides: 3 𝑎 𝑡 𝑥 − 𝑐 𝑑 𝑥 = 2 𝑎 𝑏 − 2 𝑐 3atx−cdx=2ab−2c We want to collect all terms with 𝑥 x on the right side, so add 2 𝑎 𝑏 2ab to both sides: 𝑥 = 2 𝑎 𝑏 − 2 𝑐 3 𝑎 𝑡 − 𝑐 𝑑 x= 3at−cd 2ab−2c ​ Question 3 Rearrange this expression into quadratic form 𝑎 𝑥 2 + 𝑏 𝑥 + 𝑐 = 0 ax 2 +bx+c=0 and identify the values of 𝑎 a, 𝑏 b, and 𝑐 c. 0.20 = 𝑥 75 − 𝑥 1 0.20= 75 x ​ − 1 x ​ Answer The equation is given; arrange it as follows: 0.20 = 𝑥 75 − 𝑥 0.20= 75 x ​ −x 0.20 = 𝑥 ( 1 75 − 1 ) 0.20=x( 75 1 ​ −1) 75 𝑥 + 𝑥 = 15 75x+x=15 Compare the equation 𝑥 2 − 75 𝑥 − 15 = 0 x 2 −75x−15=0 with 𝑎 𝑥 2 + 𝑏 𝑥 − 𝑐 = 0 ax 2 +bx−c=0 and write values of 𝑎 a, 𝑏 b, and 𝑐 c: 𝑎 = 75 , 𝑏 = 1 ,  and  𝑐 = − 15. a=75,b=1, and c=−15. Question 4 Determine whether the equation is always, sometimes, or never true: 2 3 𝑥 + 4 = 2 𝑥 3x+4 2 ​ =2x Answer Subtract  3 2  from both sides:  Subtract 3 2  from both sides:  𝑥 2 = 4.00. x 2 =4.00. It appears part of the explanation text was cut off, let me correct it. Here's the clean continuation for Question 4: Question 4 (Continued) Given the equation: 2 3 𝑥 + 4 = 2 𝑥 3x+4 2 ​ =2x Simplify and Solve: Multiply both sides by 3 𝑥 + 4 3x+4 (assuming 3 𝑥 + 4 ≠ 0 3x+4  =0): 2 = 2 𝑥 ( 3 𝑥 + 4 ) 2=2x(3x+4) Expand the right-hand side: 2 = 6 𝑥 2 + 8 𝑥 2=6x 2 +8x Rearrange to form a standard quadratic equation: 6 𝑥 2 + 8 𝑥 − 2 = 0 6x 2 +8x−2=0 Divide through by 2 for simplicity: 3 𝑥 2 + 4 𝑥 − 1 = 0 3x 2 +4x−1=0 Solve: Using the quadratic formula: 𝑥 = − 𝑏 ± 𝑏 2 − 4 𝑎 𝑐 2 𝑎 , 𝑎 = 3 ,   𝑏 = 4 ,   𝑐 = − 1 x= 2a −b± b 2 −4ac ​ ​ ,a=3,b=4,c=−1 𝑥 = − 4 ± 4 2 − 4 ( 3 ) ( − 1 ) 2 ( 3 ) x= 2(3) −4± 4 2 −4(3)(−1) ​ ​ 𝑥 = − 4 ± 16 + 12 6 x= 6 −4± 16+12 ​ ​ 𝑥 = − 4 ± 28 6 x= 6 −4± 28 ​ ​ 𝑥 = − 4 ± 2 7 6 x= 6 −4±2 7 ​ ​ 𝑥 = − 2 ± 7 3 x= 3 −2± 7 ​ ​ Thus, the equation is sometimes true, for the two solutions 𝑥 = − 2 + 7 3 x= 3 −2+ 7 ​ ​ or 𝑥 = − 2 − 7 3 x= 3 −2− 7 ​ ​ . Question 5 What is the solution to the following system of equations? 3 𝑥 − 6 𝑦 = − 12 3x−6y=−12 𝑥 − 2 𝑦 = − 8 x−2y=−8 (a) Use the substitution method to justify that the given system of equations has no solution. (b) What do you know about the two lines in this system of equations? Answer Part (a) The given system of equations: 3 𝑥 − 6 𝑦 = − 12 3x−6y=−12 𝑥 − 2 𝑦 = − 8 x−2y=−8 From Equation (2): 𝑥 = 2 𝑦 − 8 x=2y−8 Substitute 𝑥 = 2 𝑦 − 8 x=2y−8 into Equation (1): 3 ( 2 𝑦 − 8 ) − 6 𝑦 = − 12 3(2y−8)−6y=−12 Simplify: 6 𝑦 − 24 − 6 𝑦 = − 12 6y−24−6y=−12 − 24 = − 12 −24=−12 This is a contradiction, so there is no solution. Part (b) Look at the system of equations again: 3 𝑥 − 6 𝑦 = − 12 3x−6y=−12 𝑥 − 2 𝑦 = − 8 x−2y=−8 Rewrite both equations in slope-intercept form: For Equation (1): 3 𝑥 − 6 𝑦 = − 12 3x−6y=−12 𝑦 = 1 2 𝑥 + 2 y= 2 1 ​ x+2 For Equation (2): 𝑥 − 2 𝑦 = − 8 x−2y=−8 𝑦 = 1 2 𝑥 + 4 y= 2 1 ​ x+4 Both lines have the same slope ( 𝑚 = 1 2 m= 2 1 ​ ) but different y-intercepts ( 𝑏 = 2 b=2 and 𝑏 = 4 b=4). Thus, the two lines are parallel and will never intersect. Hence, there is no solution.