Hi. I have submitted your work. It is in doc format due to Studypool’s quality requirements. If there is any room for improvement, please do let me know. Have a great day. Section 1.1:Answer 1:Given:π¦ β² = 3π₯ 2 _______(1)π¦ = π₯ 3 + 7 _______(2)Here we need to prove the solution of a given differential equation using substitution.Equation (1) can also be written as:ππ¦= 3π₯ 2ππ₯Substitute Equation (2) to Equation (1), we get:π(π₯ 3 + 7)= 3π₯ 2ππ₯Solving the L.H.S further:π(π₯ 3 ) π(7)+= 3π₯ 2ππ₯ππ₯3π₯ 2 = 3π₯ 2Therefore L.H.S. = R.H.S.Hence Proved.Answer 19:Given:π¦ β² = π¦ + 1 _______(1)π¦(π₯) = πΆπ π₯ β 1 _______(2)π¦(0) = 5 _______(3)Calculating the derivative of Equation (2):π¦ β² (π₯) = πΆπ π₯ _______(4)Substituting Equation (2) into Equation (1):π¦ β² = πΆπ π₯ β 1 + 1π¦ β² = πΆπ π₯This returns the value of Equation (4) which was the derivative of Equation (2)Now Solving for the value of C, substitute value of π¦(0) to Equation (2):π¦(0) = πΆπ 0 β 15=πΆβ1πΆ =5+1πΆ=6The given initial condition is:π(π) = πππ β πSection 1.2:Answer 2:Given:ππ¦= (π₯ β 2)2ππ₯π¦(2) = 1In order to find the required equation w.r.t the given initial condition, we need to take the antiderivative of the above differential equation:ππ¦= (π₯ β 2)2ππ₯ππ¦ = (π₯ β 2)2 ππ₯β« ππ¦ = β«(π₯ β 2)2 ππ₯π¦=(π₯ β 2)3+πΆ3The value of C is unknown. Plugging in the value of π¦(2) = 1 to find the value of C.(2 β 2)3π¦(2) =+πΆ31= 0+πΆπΆ=1The solution is:(π β π)ππ=+ππAnswer 13:Given:π(π‘) = 3π‘π£0 = 5π₯0 = 0In order to find the velocity function, we will integrate the acceleration function w.r.t to βtβ:β« π(π‘) = β« 3π‘β«ππ£= β« 3π‘ππ‘β« ππ£ = 3. β« π‘ ππ‘π£=3π‘ 2+ πΆ12Plugging in the value of initial condition π£(0) = 5 in above equation. Note that v0 stands forvelocity at initial time (t=0)3(0)25=+ πΆ12πΆ1 = 5Therefore, the velocity function is:ππππ(π) =+ππTaking the anti-derivative again to find the displacement function:β«ππ₯3π‘ 2= β«+5ππ‘23π‘ 2β« ππ₯ = β« (+ 5) ππ‘23 π‘3π₯ = . + 5π‘ + πΆ22 3π‘3π₯ = + 5π‘ + πΆ22Plugging in the value of π₯0 = 0 in above equation to find the initial value equation0=(0)3+ 5(0) + πΆ22πΆ2 = 0Therefore:π(π) =ππ+ πππAnswer 28:Given:Initial velocity = π£0 = 40 ππ‘/π Height of Monument = π¦(π‘) = 555 ππ‘πππ‘Acceleration = π = 9.81 π 2 = 32 π 2We know that the second equation of motion is:1βπ₯ = π£0 π‘ + ππ‘ 2 ______(1)2Plugging in the values in above equation;1555 = (40)π‘ + (32)π‘ 2216π‘ 2 + 40π‘ β 555 = 0Taking the roots of above equation:π‘=β40 Β± β402 + 4π₯16π₯5552π₯16ππ; π‘ = 4.78 π ππPerforming the derivation on the equation (1) w.r.t βtβ to obtain velocity function:π£(π‘) = π£0 + ππ‘Plugging in the values:π£(π‘) = 40 + 32(4.77)π(π) = πππ. ππ ππ/πHence this will be the final velocity of the ball at impact.Section 1.4:Answer 2:Given:ππ¦+ 2π₯π¦ 2 = 0ππ₯In order to find the general solution, it is important to separate the two variables x and y andthen perform integration:ππ¦= β2π₯π¦ 2ππ₯ππ¦ = β2π₯π¦ 2 ππ₯1ππ¦ = β2π₯ ππ₯π¦2β«1ππ¦ = β2 β« π₯ ππ₯π¦2β« π¦ β2 ππ¦ = β2 β« π₯ ππ₯π¦ β1 β2π₯ 2=+πΆβ12β1= βπ₯ 2 + πΆπ¦1= π₯2 β πΆπ¦π=πππβπͺAnswer 11:Given:π¦ β² = π₯π¦ 3ππ,ππ¦= π₯π¦ 3ππ₯In order to find the general solution, we will separate the two variables x and y and performtheir integration separately:ππ¦ = π₯π¦ 3 ππ₯1ππ¦ = …

Next to these stages, there are four types of loyalty. The first type is no loyalty: this can be seen as the typical action supporter that is giving money to fundraiser friends/family. There is no connection between the action supporter and War Child, non-loyal customers add a small amount to the financial account of the organization. The second type is inertia loyalty: this type can be a potential Friend, the person is giving a donation because they always have done it. This person is not loyal to one charity because there is no deeper connection and is easy with switching from charity to charity. This person is reachable to get a deeper connection with to become a potential Friend. It is needed to give the potential Friend attention and please the person. The third type is latent loyalty: this type is based on situational influences such as time, social life, physical factors, reasons why to support charity and the mood of the person. Latent loyalty means that there are repeated purchases. The last type is premium loyalty: this is the greatest type an organization can get. This type indicates the loyal Friend who is proud to discover more and tell about War Child to everyone. (Griffin, 2010) Using the stages and types of loyalty, War Child will get a better understanding about their potential Friends. Increased loyalty can give benefits to more respects of the company. Increased loyalty will be cost saving because there will be reduced marketing costs, more positive word-of-mouth, satisfied loyal customers whereby the failure costs decrease. (Griffin, 2010) 1.2.4 Customer relationship management Kotler (2014, p. 9) stated in the book ββPrinciples of marketingββ that customer relationship management (CRM) can be mentioned as delivering superior customer value and satisfaction by building and maintaining profitable customer relationships. (Kotler, 2014) Using a suitable approach for the action supporters will lead to charitable giving. To build and maintain profitable customer relationships, customer relationship management need to be implemented. It is very important to deliver customer value and make the action supporters satisfied. By doing this,>

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