# SOLUTION: Direct Integration Separation of Variables & Integrating Factor Method Questions

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¦ β² = 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:ππ¦= 3π₯ 2ππ₯Substitute Equation (2) to Equation (1), we get:π(π₯ 3 + 7)= 3π₯ 2ππ₯Solving the L.H.S further:π(π₯ 3 ) π(7)+= 3π₯ 2ππ₯ππ₯3π₯ 2 = 3π₯ 2Therefore L.H.S. = R.H.S.Hence Proved.Answer 19:Given:π¦ β² = π¦ + 1 _______(1)π¦(π₯) = πΆπ π₯ β 1 _______(2)π¦(0) = 5 _______(3)Calculating the derivative of Equation (2):π¦ β² (π₯) = πΆπ π₯ _______(4)Substituting Equation (2) into Equation (1):π¦ β² = πΆπ π₯ β 1 + 1π¦ β² = πΆπ 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) = πΆπ 0 β 15=πΆβ1πΆ =5+1πΆ=6The given initial condition is:π(π) = πππ β πSection 1.2:Answer 2:Given:ππ¦= (π₯ β 2)2ππ₯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:ππ¦= (π₯ β 2)2ππ₯ππ¦ = (π₯ β 2)2 ππ₯β« ππ¦ = β«(π₯ β 2)2 ππ₯π¦=(π₯ β 2)3+πΆ3The value of C is unknown. Plugging in the value of π¦(2) = 1 to find the value of C.(2 β 2)3π¦(2) =+πΆ31= 0+πΆπΆ=1The solution is:(π β π)ππ=+ππAnswer 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 βtβ:β« π(π‘) = β« 3π‘β«ππ£= β« 3π‘ππ‘β« ππ£ = 3. β« π‘ ππ‘π£=3π‘ 2+ πΆ12Plugging 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=+ πΆ12πΆ1 = 5Therefore, the velocity function is:ππππ(π) =+ππTaking the anti-derivative again to find the displacement function:β«ππ₯3π‘ 2= β«+5ππ‘23π‘ 2β« ππ₯ = β« (+ 5) ππ‘23 π‘3π₯ = . + 5π‘ + πΆ22 3π‘3π₯ = + 5π‘ + πΆ22Plugging in the value of π₯0 = 0 in above equation to find the initial value equation0=(0)3+ 5(0) + πΆ22πΆ2 = 0Therefore:π(π) =ππ+ πππAnswer 28:Given:Initial velocity = π£0 = 40 ππ‘/π Height of Monument = π¦(π‘) = 555 ππ‘πππ‘Acceleration = π = 9.81 π 2 = 32 π 2We know that the second equation of motion is:1βπ₯ = π£0 π‘ + ππ‘ 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‘=β40 Β± β402 + 4π₯16π₯5552π₯16ππ; π‘ = 4.78 π ππPerforming the derivation on the equation (1) w.r.t βtβ to obtain velocity function:π£(π‘) = π£0 + ππ‘Plugging in the values:π£(π‘) = 40 + 32(4.77)π(π) = πππ. ππ ππ/πHence this will be the final velocity of the ball at impact.Section 1.4:Answer 2:Given:ππ¦+ 2π₯π¦ 2 = 0ππ₯In order to find the general solution, it is important to separate the two variables x and y andthen perform integration:ππ¦= β2π₯π¦ 2ππ₯ππ¦ = β2π₯π¦ 2 ππ₯1ππ¦ = β2π₯ ππ₯π¦2β«1ππ¦ = β2 β« π₯ ππ₯π¦2β« π¦ β2 ππ¦ = β2 β« π₯ ππ₯π¦ β1 β2π₯ 2=+πΆβ12β1= βπ₯ 2 + πΆπ¦1= π₯2 β πΆπ¦π=πππβπͺAnswer 11:Given:π¦ β² = π₯π¦ 3ππ,ππ¦= π₯π¦ 3ππ₯In order to find the general solution, we will separate the two variables x and y and performtheir integration separately:ππ¦ = π₯π¦ 3 ππ₯1ππ¦ = …

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