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 ′ = 3𝑥 2 _______(1)𝑦 = 𝑥 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= 3𝑥 2𝑑𝑥Substitute Equation (2) to Equation (1), we get:𝑑(𝑥 3 + 7)= 3𝑥 2𝑑𝑥Solving the L.H.S further:𝑑(𝑥 3 ) 𝑑(7)+= 3𝑥 2𝑑𝑥𝑑𝑥3𝑥 2 = 3𝑥 2Therefore L.H.S. = R.H.S.Hence Proved.Answer 19:Given:𝑦 ′ = 𝑦 + 1 _______(1)𝑦(𝑥) = 𝐶𝑒 𝑥 − 1 _______(2)𝑦(0) = 5 _______(3)Calculating the derivative of Equation (2):𝑦 ′ (𝑥) = 𝐶𝑒 𝑥 _______(4)Substituting Equation (2) into Equation (1):𝑦 ′ = 𝐶𝑒 𝑥 − 1 + 1𝑦 ′ = 𝐶𝑒 𝑥This 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) = 𝐶𝑒 0 − 15=𝐶−1𝐶 =5+1𝐶=6The given initial condition is:𝒚(𝒙) = 𝟔𝒆𝒙 − 𝟏Section 1.2:Answer 2:Given:𝑑𝑦= (𝑥 − 2)2𝑑𝑥𝑦(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= (𝑥 − 2)2𝑑𝑥𝑑𝑦 = (𝑥 − 2)2 𝑑𝑥∫ 𝑑𝑦 = ∫(𝑥 − 2)2 𝑑𝑥𝑦=(𝑥 − 2)3+𝐶3The value of C is unknown. Plugging in the value of 𝑦(2) = 1 to find the value of C.(2 − 2)3𝑦(2) =+𝐶31= 0+𝐶𝐶=1The solution is:(𝒙 − 𝟐)𝟑𝒚=+𝟏𝟑Answer 13:Given:𝑎(𝑡) = 3𝑡𝑣0 = 5𝑥0 = 0In order to find the velocity function, we will integrate the acceleration function w.r.t to ‘t’:∫ 𝑎(𝑡) = ∫ 3𝑡∫𝑑𝑣= ∫ 3𝑡𝑑𝑡∫ 𝑑𝑣 = 3. ∫ 𝑡 𝑑𝑡𝑣=3𝑡 2+ 𝐶12Plugging 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=+ 𝐶12𝐶1 = 5Therefore, the velocity function is:𝟑𝒕𝟐𝒗(𝒕) =+𝟓𝟐Taking the anti-derivative again to find the displacement function:∫𝑑𝑥3𝑡 2= ∫+5𝑑𝑡23𝑡 2∫ 𝑑𝑥 = ∫ (+ 5) 𝑑𝑡23 𝑡3𝑥 = . + 5𝑡 + 𝐶22 3𝑡3𝑥 = + 5𝑡 + 𝐶22Plugging in the value of 𝑥0 = 0 in above equation to find the initial value equation0=(0)3+ 5(0) + 𝐶22𝐶2 = 0Therefore:𝒙(𝒕) =𝒕𝟑+ 𝟓𝒕𝟐Answer 28:Given:Initial velocity = 𝑣0 = 40 𝑓𝑡/𝑠Height of Monument = 𝑦(𝑡) = 555 𝑓𝑡𝑚𝑓𝑡Acceleration = 𝑎 = 9.81 𝑠2 = 32 𝑠2We know that the second equation of motion is:1∆𝑥 = 𝑣0 𝑡 + 𝑎𝑡 2 ______(1)2Plugging in the values in above equation;1555 = (40)𝑡 + (32)𝑡 2216𝑡 2 + 40𝑡 − 555 = 0Taking the roots of above equation:𝑡=−40 ± √402 + 4𝑥16𝑥5552𝑥16𝑜𝑟; 𝑡 = 4.78 𝑠𝑒𝑐Performing the derivation on the equation (1) w.r.t ‘t’ to obtain velocity function:𝑣(𝑡) = 𝑣0 + 𝑎𝑡Plugging in the values:𝑣(𝑡) = 40 + 32(4.77)𝒗(𝒕) = 𝟏𝟗𝟐. 𝟔𝟖 𝒇𝒕/𝒔Hence this will be the final velocity of the ball at impact.Section 1.4:Answer 2:Given:𝑑𝑦+ 2𝑥𝑦 2 = 0𝑑𝑥In order to find the general solution, it is important to separate the two variables x and y andthen perform integration:𝑑𝑦= −2𝑥𝑦 2𝑑𝑥𝑑𝑦 = −2𝑥𝑦 2 𝑑𝑥1𝑑𝑦 = −2𝑥 𝑑𝑥𝑦2∫1𝑑𝑦 = −2 ∫ 𝑥 𝑑𝑥𝑦2∫ 𝑦 −2 𝑑𝑦 = −2 ∫ 𝑥 𝑑𝑥𝑦 −1 −2𝑥 2=+𝐶−12−1= −𝑥 2 + 𝐶𝑦1= 𝑥2 − 𝐶𝑦𝒚=𝒙𝟐𝟏−𝑪Answer 11:Given:𝑦 ′ = 𝑥𝑦 3𝑜𝑟,𝑑𝑦= 𝑥𝑦 3𝑑𝑥In order to find the general solution, we will separate the two variables x and y and performtheir integration separately:𝑑𝑦 = 𝑥𝑦 3 𝑑𝑥1𝑑𝑦 = …
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