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:𝑦 β€² = 3π‘₯ 2 _______(1)𝑦 = 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:𝑑𝑦= 3π‘₯ 2𝑑π‘₯Substitute Equation (2) to Equation (1), we get:𝑑(π‘₯ 3 + 7)= 3π‘₯ 2𝑑π‘₯Solving the L.H.S further:𝑑(π‘₯ 3 ) 𝑑(7)+= 3π‘₯ 2𝑑π‘₯𝑑π‘₯3π‘₯ 2 = 3π‘₯ 2Therefore L.H.S. = R.H.S.Hence Proved.Answer 19:Given:𝑦 β€² = 𝑦 + 1 _______(1)𝑦(π‘₯) = 𝐢𝑒 π‘₯ βˆ’ 1 _______(2)𝑦(0) = 5 _______(3)Calculating the derivative of Equation (2):𝑦 β€² (π‘₯) = 𝐢𝑒 π‘₯ _______(4)Substituting Equation (2) into Equation (1):𝑦 β€² = 𝐢𝑒 π‘₯ βˆ’ 1 + 1𝑦 β€² = 𝐢𝑒 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 𝑦(0) to Equation (2):𝑦(0) = 𝐢𝑒 0 βˆ’ 15=πΆβˆ’1𝐢 =5+1𝐢=6The given initial condition is:π’š(𝒙) = πŸ”π’†π’™ βˆ’ 𝟏Section 1.2:Answer 2:Given:𝑑𝑦= (π‘₯ βˆ’ 2)2𝑑π‘₯𝑦(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:𝑑𝑦= (π‘₯ βˆ’ 2)2𝑑π‘₯𝑑𝑦 = (π‘₯ βˆ’ 2)2 𝑑π‘₯∫ 𝑑𝑦 = ∫(π‘₯ βˆ’ 2)2 𝑑π‘₯𝑦=(π‘₯ βˆ’ 2)3+𝐢3The value of C is unknown. Plugging in the value of 𝑦(2) = 1 to find the value of C.(2 βˆ’ 2)3𝑦(2) =+𝐢31= 0+𝐢𝐢=1The solution is:(𝒙 βˆ’ 𝟐)πŸ‘π’š=+πŸπŸ‘Answer 13:Given:π‘Ž(𝑑) = 3𝑑𝑣0 = 5π‘₯0 = 0In order to find the velocity function, we will integrate the acceleration function w.r.t to β€˜t’:∫ π‘Ž(𝑑) = ∫ 3π‘‘βˆ«π‘‘π‘£= ∫ 3π‘‘π‘‘π‘‘βˆ« 𝑑𝑣 = 3. ∫ 𝑑 𝑑𝑑𝑣=3𝑑 2+ 𝐢12Plugging in the value of initial condition 𝑣(0) = 5 in above equation. Note that v0 stands forvelocity at initial time (t=0)3(0)25=+ 𝐢12𝐢1 = 5Therefore, the velocity function is:πŸ‘π’•πŸπ’—(𝒕) =+πŸ“πŸTaking the anti-derivative again to find the displacement function:βˆ«π‘‘π‘₯3𝑑 2= ∫+5𝑑𝑑23𝑑 2∫ 𝑑π‘₯ = ∫ (+ 5) 𝑑𝑑23 𝑑3π‘₯ = . + 5𝑑 + 𝐢22 3𝑑3π‘₯ = + 5𝑑 + 𝐢22Plugging in the value of π‘₯0 = 0 in above equation to find the initial value equation0=(0)3+ 5(0) + 𝐢22𝐢2 = 0Therefore:𝒙(𝒕) =π’•πŸ‘+ πŸ“π’•πŸAnswer 28:Given:Initial velocity = 𝑣0 = 40 𝑓𝑑/𝑠Height of Monument = 𝑦(𝑑) = 555 π‘“π‘‘π‘šπ‘“π‘‘Acceleration = π‘Ž = 9.81 𝑠2 = 32 𝑠2We know that the second equation of motion is:1βˆ†π‘₯ = 𝑣0 𝑑 + π‘Žπ‘‘ 2 ______(1)2Plugging in the values in above equation;1555 = (40)𝑑 + (32)𝑑 2216𝑑 2 + 40𝑑 βˆ’ 555 = 0Taking the roots of above equation:𝑑=βˆ’40 Β± √402 + 4π‘₯16π‘₯5552π‘₯16π‘œπ‘Ÿ; 𝑑 = 4.78 𝑠𝑒𝑐Performing the derivation on the equation (1) w.r.t β€˜t’ to obtain velocity function:𝑣(𝑑) = 𝑣0 + π‘Žπ‘‘Plugging in the values:𝑣(𝑑) = 40 + 32(4.77)𝒗(𝒕) = πŸπŸ—πŸ. πŸ”πŸ– 𝒇𝒕/𝒔Hence this will be the final velocity of the ball at impact.Section 1.4:Answer 2:Given:𝑑𝑦+ 2π‘₯𝑦 2 = 0𝑑π‘₯In order to find the general solution, it is important to separate the two variables x and y andthen perform integration:𝑑𝑦= βˆ’2π‘₯𝑦 2𝑑π‘₯𝑑𝑦 = βˆ’2π‘₯𝑦 2 𝑑π‘₯1𝑑𝑦 = βˆ’2π‘₯ 𝑑π‘₯𝑦2∫1𝑑𝑦 = βˆ’2 ∫ π‘₯ 𝑑π‘₯𝑦2∫ 𝑦 βˆ’2 𝑑𝑦 = βˆ’2 ∫ π‘₯ 𝑑π‘₯𝑦 βˆ’1 βˆ’2π‘₯ 2=+πΆβˆ’12βˆ’1= βˆ’π‘₯ 2 + 𝐢𝑦1= π‘₯2 βˆ’ πΆπ‘¦π’š=π’™πŸπŸβˆ’π‘ͺAnswer 11:Given:𝑦 β€² = π‘₯𝑦 3π‘œπ‘Ÿ,𝑑𝑦= π‘₯𝑦 3𝑑π‘₯In order to find the general solution, we will separate the two variables x and y and performtheir integration separately:𝑑𝑦 = π‘₯𝑦 3 𝑑π‘₯1𝑑𝑦 = …

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