Worked Solutions 27 Chapter 6: Simplifying Through Substitution 6 a Let u = 3 + 3y + 2 Solving for y computing y we then see that 3y = u 3 2 y = u 3 2 3 = d [ u 3 2 3 = 3 Substituting this into the differential equation solving for u we have = 3 + 3y + 2 3 = 3 = + = + u2 + = 3 + = = + u2 3 = 3 + c ) + u2 Computing the last integral ignoring the arbitrary constant): + u + = 2 + [ = = u arctanu) + So equation ) becomes u arctanu) = 3 + c which cannot be solved for u So we now replace each u with its formula in terms of y from the original substitution u = 3 + 3y + 2 ) simplify the resulting equation as much as practical: 3 + 3y + 2 arctan3 + 3y + 2) = 3 + c arctan3 + 3y + 2) = 3y + 2 + c = 3y + C 3 + 3y + 2 = tan3y + C) 6 c Using the substitution u = 4y 8 + 3 we have y = 4 [u + 8 3 = u 4 + 2 3 4 = 4 + 2 cos4y 8 + 3) = 2 + 2 cos4y 8 + 3) [ cosu) 4 + 2 = 2 + 2 cosu)
28 Simplifying Through Substitution cosu) 4 + 2 cosu) = 2 + 2 cosu) cosu) = 8 cosu) = 8 sinu) = 8 + c u = arcsin8 + c) Plugging the last formula for u into the formula for y based on the original substitution yields the final answer y = [u + 8 3 4 = 4 [arcsin8 + c) + 8 3 = 4 arcsin8 + c) + 2 3 4 63 a Rewriting the differential equation: 2 y = y2 = y y + Since the equation is homogeneous our substitution is based on u = y from which we derive y = u = d Using this with our differential equation: [u = u + = u + = y y + u + = u + u2 Obviously u = 0 is the only constant solution which in turn also yields = u2 y = u = 0 = 0 as a constant solution to the original equation For the other solutions: = u2 u 2 = u 2 = u = ln + c u = ln + c This along with our formula for y in terms of u yields [ y = u = = ln + c ln + c which along with y = 0 describes all the solutions 63 c The substitution: u = y y = u = d [u = u +
Worked Solutions 29 So y )[ cos [ cosu) y u + ) u y ) = + sin = + sinu) cosu) = + sinu) The constant solutions of this are given by those values of u where + sinu) = 0 ; that is u = arcsin ) = 2nπ π for n = 0 ± ±2 2 For the nonconstant solutions: = + sinu) cosu) cosu) + sinu) = cosu) + sinu) = ln + sinu) = ln + c + sinu) = ±e ln +c = A u = arcsina ) Note that the last formula for u also yields the constant solutions So all the solutions to our original differential equation are given by y = u = arcsina ) 65 a First observe that y = 0 is clearly a constant solution Here n = 3 from the 3y 3 term) So the substitution is based on u = y 3 = y 2 Hence we also have y = ±u / 2 Using this substitution: = d [±u / 2 [ = ± 2 u 3 / 2 + 3y = 3y3 ) 2 u 3 / 2 3 + 3u / 2 = 3 u / 2 2u 3 )[ / 2 2 u 3 / 2 + 3u / 2 = 3u 3 / 2 6u = 6 The integrating factor for the last differential equation is µ) = e 6) = e 6 Using this with the last differential equation above: [ e 6 6u = 6 e 6 6e 6 u = 6e 6 d [ e 6 u = 6e 6 e 6 u = 6e 6 = e 6 + c u = + ce 6
30 Simplifying Through Substitution This with the original substitution then gives y = ±u / 2 = ± + ce 6) / 2 So any solution to the original differential equation is given by the above formula for y or y = 0 65 c Again we note that y = 0 is a solution The substitution: u = y 2 / 3 = y / 3 y = u 3 Using the substitution to find the linear equation for u : + 3 cot)y = 6 cos)y2 / 3 3 + 3 cot)u3 = 6 cos) u 3)2 / 3 + cot)u = 2 cos) = 3u2 The integrating factor is given by ) µ) = cot) cos) e = ep sin) = epln sin) ) = sin) As noted in chapter 5 we can simply use µ = sin) Doing so: [ sin) + cot)u = 2 cos) d [sin)u = 2 sin) cos) sin)u = 2 sin) cos) = sin 2 ) + c u = sin) + c sin) y = u 3 = sin) + c ) 3 sin) The last equation along with y = 0 describes all solutions to our original differential equation 67 a This is obviously in the form = formula of y So it is a homogeneous equation we use the corresponding substitution u = y y = u = d [u = u +
Worked Solutions 3 Doing so: = y + u + y = u + u 2 = = 3 u3 = ln + c u = 3 ln + 3c) / 3 y = u = 3 ln + C) / 3 67 c Since the equation is + 2 y = 4y / 2 it is a Bernoulli equation with n = / 2 One solution is y = 0 The others are obtained using the substitution Hence u = y / 2 = y / 2 y = = 2u + 2 y = 4y / 2 2u + 2 u2 = 4 ) / 2 + u = 2 The integrating factor is µ) = e / = = So + u = 2 [ + u = 2 d [u = 2 u = + c u = y = = 2 = 2 + c + c 67 e The y factor suggests using the linear substitution u = y Then Using this y = u + = d y ) = u [ + u = u [u + = + = u u Note that the derivative of u is zero if only if u = So u = is the only constant solution to this last equation The corresponding solution to the original equation is then y = u + = + =
32 Simplifying Through Substitution For the other solutions we ll need u u u = u [ = u Back to the last differential equation above: u u = = u u = u ln u + C u u = u ln u = + c y ) ln y ) = + c ln y + = c y y + = ±e c y = Ae y y = + Ae y This last equation describes all solutions since it does rece to the solution y) = + when A = 0 67 g = 2 + 2y + 2y 2 2y + 2 2 = 2y + 2 2) = 2 + 2y + 2y 2 [ 2 + 2 y y ) 2 + 2 [ 2 2 y = y y + 2 + 2 + 2 2 y + 2 So the equation is homogeneous we use the substitution u = y Hence y = u = d [u = u + = + 2 y y + 2 2 y u + + 2 + u2 = ) = u = + u2 + u2 = 2u2 + 2u = ) = + 2u = ln + c + 2u ln + c) = 0
Worked Solutions 33 u = 2 ± 2 2 4[ ln + c) = ± ln + C 2 [ y = u = ± ln + C = ± ln + C 67 i Clearly the substitution u = 2 + y 3 is in order With this y = u 2 + 3 = d [u 2 + 3 = 2 = 2 2 + y 3 2 2 = 2 u 2 = 2 u Note that this last equation has one constant solution u = 0 Corresponding to this is the solution to the original equation y = u 2 + 3 = 0 2 + 3 = 3 2 For the other solutions we continue the above computations: = u u / 2 = 2 2u / 2 = 2 + C u = + c y = u 2 + 3 = + c 2 + 3 So each solution is given by either the last formula or by y = 3 2 67 k y = y + 2 = y + y + 2 = y + y + )/ 2 So the equation is homogeneous we use the substitution u = y Hence y = u = d [u = u + = y + y + )/ 2 u + = u + u + ) / 2 = u + ) / 2 ) Note that the derivative of u is zero if u = So u = is a constant solution the corresponding solution to our original equation is y = u = ) =
34 Simplifying Through Substitution For the rest of the solutions we divide equation ) by u + ) / 2 continue the process: u + ) / 2 = u + ) / 2 = 2u + ) / 2 = ln + C u = y = u = [ 2 ln + c = 2 ln + c 2 ln + c So every solution is given by either the last formula for y or by y = 67 m Clearly the linear substitution u = y + 3 is appropriate Using this y = u + 3 = d [ u + 3 = = y + 3)2 = u2 = u2 Here the derivative of u is zero if u = ± So this differential equation has two constant solutions u = u = Since y = u + 3 the corresponding solutions to the original equation are y = + 3 = + 2 y = ) + 3 = + 4 To find the other solutions we divide the last differential equation above by proceed as usual using partial fractions to rewrite ) in more convenient form for integration: = = + u) u) = + c [ 2 + u + = + c u [ln + u ln u = + c 2 ln + u = 2 + 2c u + u u = ±e2s+2c = Ae 2 + u = Ae 2 u Ae 2 u = Ae2 Ae 2 + y) = u + 3 = + 3 + Ae2 Ae 2 +
Worked Solutions 35 Note that the last formula for y reces to the constant solution y = + 2 if A = 0 but does not rece to the constant solution y = + 4 for any value of A So to describe all solutions we need both y = + 3 + Ae2 Ae 2 + y = + 4 67 o We need a formula of y maybe f y) such that the differential equation reces to a simple differential equation for u when we let u = f y) Because the given equation involves both siny) cosy) two possible choices for the substitution are obviously Differentiating these yield respectively u = cosy) u = siny) = siny) = cosy) the second of which matches eactly the left side of the given differential equation So let s use the second choice u = siny) as our substitution Doing so we get with = cosy) cosy) = e siny) = e u + u = e This is a simple linear equation with integrating factor µ) = e = e Multiplying our differential equation for u by this integrating factor continuing: e u = [ e + u = e d [ e u = = + c u = + c)e siny) = + c)e y = arcsin + c)e )