■ Good to understand where it comes from, but we can remember this as a rule: •When you come across such an equation of the form dy dx + P(x)Y = 9(x)
■ It is analytically soluble, simply find the integrating factor then evaluate
tg,y-cigm
= of p(x)dx = µq (x) dx
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INTEGRATING FACTOR EXERCISE
■An RL circuit with a ramp input behaves according to the following differential equation for current, i di iR + L — = St dt
• Find a solution for i as a function of t given i(0) = 0
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FACTOR EXERCISE SOLN (1)
di iR + L — = 5t dt ■ Recognise and put into standard form di L — + iR = 5t dt di R5t + =7 ■This is the standard form with dtR + p(t)i = q(t) : p(t) q(t) = Gt
,,rohriggirc
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INTEGRATING FACTOR EXERCISE SOLN (2) dt di + p(t)i = q(t) • p(t) = —R q(t) = Gt Apply integrating factor =eIp(t) dt = e4dt
Evaluate the integral
f dt = t
�(�=eLt
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INTEGRATING FACTOR EXERCISE SOLN (3) RL
Put this into our result /Ai = pq(t)dt eR = f (t,)dt . .
INTEGRATING FACTOR EXERCISE SOLN (3) Rt µ—eL Put this into our result Pi = 1 pq(t)dt
Rt = 1 At(t,) dt
Solve — using parts…
Note — can’t just integrate directly, all the rules we learnt in integration still apply, we need parts’,
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INTEGRATING FACTOR EXERCISE SOLN (4) RC eLi=51 q 17, to dt u = t ; dv = dt —> du=dt;v=ReT — J ReTdt} Rt eLi= T14 e14 L2 T2 e14 + Note — Constant of integration really important here!!
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INTEGRATING FACTOR EXERCISE SOLN (5) (Lt eR: 702 eRLt c}
INTEGRATING FACTOR EXERCISE SOLN (5) eRtti f 17/,,2 eR: + Rt 5t 5L SC eLi= iteL (5t 5L Sc Sc) I=V –Tj+Te To find C – we must apply an initial condition e.g. i(0) = 0 47,Fxr, 5L SC L2 0= + C = 22/04/2018 – ENG01001 13 of 35 INTEGRATING FACTOR EXERCISE SOLN (6) = f5t 5L 5C Rt} —R – —R2 + —Le L
And So
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L2 C 1:7 St 5L 5L = _Rt} f-R – —R2 + —R2e L
i = 51-t–R R2 – 1)1
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RECALL – DEFINITIONS (II)
• Inhomogeneous 2. order ODE p(x)—d2 y + (x) —dx + r (x) y = f (x) dx2 dy ■Which can also be written
RECALL HOMOGENEOUS SOLUTION
■General form ay” + by’ + cy = 0 ■We try
y = ekx
■This gives y’ = kek” y” = k(kek”) = k2e’ ■Sub into the original equation to get ak2ek” + bkek” + cek” = 0 ■The Auxiliary Equation ak2 + bk + c = 0
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HOMOGENEOUS SOLUTION
■The Auxiliary Equation ak2 +bk+c =0 ■This is a quadratic equation in the variable k —b + Vb2 — 4ac k = 2a ■ Done! With k as above, y = ek” is a solution therefore the general solution is
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yH = Aeki” + Bek2x
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HOMOGENEOUS SOLUTION IN A PAGE
ay” + by’ + cy = 0
Try solution y = ekx . ,b2,kx 1,,,,kx
^ k” = 0
RECALL – DEFINITIONS (II)
• Inhomogeneous 2. order ODE d2 y dy + q(x)-d-T, + r(x) y = f (x) •Which can also be written p(x)y” (x) + q(x)y’ (x) + r(x)y = f (x) ■Again, we will deal with constant coefficients only i.e. p (x), q(x)& r (x) are constants. ■ But note – f (x) may be a more complicated function.
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SOME DEFINITIONS (IV)
• If yp is any solution of an inhomogeneous equation then the general solution of that equation is: Y(x) = Y H (x) + Yp (x) ■Where y, (x) is the solution to the associated homogeneous equation (see slide 8) • In this situation we say —yH is the Complementary Function (C.F.) —yp is the Particular Integral (PI) (see section 7.1.2 in notes)
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RECALL HOMOGENEOUS SOLUTION
■ General form ay” + by’ + cy = 0 •We try
HOMOGENEOUS SOLUTION IN A PAGE
ay” + by’ + cy = 0
Try solution y = e” ak2ekx + bkekx + cek” = 0 Auxiliary equation ak2 + bk + c = k = b±4’2 ■ Note — as with any quadratic — in general there are two values of k. Using superposition (slide 8), usually write the solution: y = Aek, + Bek2x ■ Where k, & k2 are the two quadratic roots and A & B are constants determined by given conditions
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HOMOGENEOUS SOLUTION CASES
■As with any quadratic, we can have three cases for k5 & k2 (the roots of the Auxiliary Equation) 1. Roots are real and distinct (like the example) i.e. k, & k2 are two different real numbers 2. Roots are real and repeated i.e. k, & k2 are the same real number 3. Roots are a complex conjugate pair i.e. k, & k2 have the formk,=a+i(3&k2=a—if3 ■All three cases have general forms given in the formula sheets at the back of your notes
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AUXILIARY EQUATION FORMS
■ Distinct real roots: Auxiliary Equation yields k1, k2 as different real numbers ■ General solution is IvH = Aeklx + Bek,
AUXILIARY EQUATION FORMS
■ Distinct real roots: Auxiliary Equation yields k1, k2 as different real numbers ■ General solution is
YH = Aek, + Bek,
• Repeated root: Auxiliary Equation yields lc, = k2 • General solution is
yH = (Ax + B)ek”
■ Complex roots: Auxiliary Equation yields = a + if k2 = a — ■ General solution is
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DAMPING — THE FORM OF HOMOGENEOUS SOLUTION • The three cases described have real world meaning, they relate to damping ■ Damping is how quickly a system, when disturbed, returns to its equilibrium (stops oscillating). ■ If a system just returns to zero without oscillating, we call that system over-damped — this is the situation where we have real distinct roots i.e. where b2 > 4ac • If a system touches zero JUST as it’s about to oscillate, that is critically damped, repeated root, b2 = 4ac ■ If a system oscillates as it returns to zero, it is under damped (b2 < 4ac). Extreme: b = 0, no damping
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