Special Mathematics
Higher order linear equations
March 2018
ii
“Is there a better motivation than success ?”
Ion Tiriac
2Higher order linear equations
Bungee jumping
Bungee jumping is an activity that involves jumping from a tall structurewhile connected to a large elastic cord. The tall structure is usually a fixedobject, such as a building, bridge or crane; but it is also possible to jump froma movable object, such as a hot-air-balloon or helicopter, that has the ability tohover above the ground. The thrill comes from the free-falling and the rebound.When the person jumps, the cord stretches and the jumper flies upwards againas the cord recoils, and continues to oscillate up and down until all the kineticenergy is dissipated.
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You decided to try a bungee jump-ing. Let’s suppose that each of thecords you have brought will be tiedoff so as to be 100 feet long whenhanging from the bridge. Call the po-sition at the bottom of the cord 0,and measure the position of your feetbelow that “natural length” as 𝑥(𝑡),where 𝑥 increases as you go down andis a function of time t. See the fig-ure. Then, at the time you jump,𝑥(0) = −100, while if your six-footframe hits the water head first, at thattime 𝑥(𝑡) = 174 − 100 − 6 = 68. No-tice that distance increases as you fall,and so your velocity is positive as youfall and negative when you bounceback up.
Note also that you plan to dive so your head will be six feet below the endof the chord when it stops you. You know that the acceleration due to gravityis a constant, called 𝑔, so that the force pulling downwards on your body is𝑚𝑔. You know that when you leap from the bridge, air resistance will increaseproportionally to your speed, providing a force in the opposite direction to yourmotion of about 𝛽𝑣, where 𝛽 is a constant and 𝑣 is your velocity. Finally, youknow that Hooke’s law describing the action of springs says that the bungeecord will eventually exert a force on you proportional to its distance past itsnatural length. Thus, you know that the force of the cord pulling you back fromdestruction may be expressed as:
𝑏(𝑥) =
{︃0, 𝑥 ≤ 0
−𝑘𝑥, 𝑥 > 0
The number 𝑘 is called the spring constant, and it is where the stiffness ofthe cord you use influences the equation. For example, if you used the steelcable, then 𝑘 would be very large, giving a tremendous stopping force verysuddenly as you passed the natural length of the cable. This could lead todiscomfort, injury, or even a Darwin award. You want to choose the cord witha k value large enough to stop you above or just touching the water, but nottoo suddenly. Consequently, you are interested in finding the distance you fallbelow the natural length of the cord as a function of the spring constant. Todo that, you must solve the differential equation that we have derived in wordsabove: The force on your body is given by:
𝑚 · 𝑥′′(𝑡) = 𝑚 · 𝑔 + 𝑏(𝑥(𝑡)) − 𝛽 · 𝑥′(𝑡)
Here 𝑚𝑔 is your weight, and 𝑥 is the rate of change of your position belowthe equilibrium with respect to time: your velocity. The constant 𝛽 for air
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resistance depends on a number of things, including whether you wear yourskin-tight pink spandex or your skater shorts and XXL T-shirt, but you knowthat the value today is about 1.0.
This is a nonlinear differential equation, but inside it are two linear differ-ential equations, struggling to get out. When 𝑥 < 0, the equation is:
𝑚𝑥′′(𝑡) = 𝑚𝑔 − 𝛽𝑥′(𝑡)
while after you pass the natural length of the cord it is:
𝑚𝑥′′(𝑡) = 𝑚𝑔 − 𝑘𝑥(𝑡) − 𝛽𝑥′(𝑡)
As you can see, knowing a little bit of math is a dangerous thing. Weremind you that the assumption that the drag due to air resistance islinear applies only for low speeds. By the time you swoop past the naturallength of the cord, that approximation is only wishful thinking, so youractual mileage may vary. Moreover, springs behave nonlinearly in largeoscillations, so Hooke’s law is only an approximation. Do not trust yourlife to an approximation made by a man who has been dead for 200 years!Leave bungee jumping to the professionals.
Remark:
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Homogeneous linear equations with constant coefficients:
𝑦(𝑛) + 𝑎1𝑦(𝑛−1) + . . . + 𝑎𝑛−1𝑦
′ + 𝑎𝑛𝑦 = 0
where 𝑎𝑖 ∈ R, 𝑖 = 1, 𝑛.
Algorithm:∙ write the characteristic polynomial:
𝑝(𝜆) = 𝜆𝑛 + 𝑎1𝜆𝑛−1 . . . + 𝑎𝑛−1𝜆 + 𝑎𝑛
∙ find the roots, real or complex, of the characteristic equation: 𝑝(𝜆) = 0.∙ for every root we obtain an element of the fundamental system of solutions,
according to the following table:
root generated solution
𝜆 = 𝑎 𝑒𝑎𝑥
𝜆1 = 𝜆2 = . . . = 𝜆𝑘 = 𝑎 𝑒𝑎𝑥, 𝑥𝑒𝑎𝑥, 𝑥2𝑒𝑎𝑥. . . , 𝑥𝑘−1𝑒𝑎𝑥
𝜆 = 𝛼 + 𝛽𝑖, 𝜆 = 𝛼− 𝛽𝑖 𝑒𝛼𝑥 sin𝛽𝑥, 𝑒𝛼𝑥 cos𝛽𝑥{︃𝜆1 = 𝜆2 = . . . = 𝜆𝑘 = 𝛼 + 𝛽𝑖
𝜆𝑘+1 = 𝜆𝑘+2 = . . . = 𝜆2𝑘 = 𝛼− 𝛽𝑖
{︃𝑒𝛼𝑥 sin𝛽𝑥, 𝑥𝑒𝛼𝑥 sin𝛽𝑥 . . . , 𝑥𝑘−1𝑒𝛼𝑥 sin𝛽𝑥
𝑒𝛼𝑥 cos𝛽𝑥, 𝑥𝑒𝛼𝑥 cos𝛽𝑥 . . . , 𝑥𝑘−1𝑒𝛼𝑥 cos𝛽𝑥
𝜆 = 0 1
𝜆1 = 𝜆2 = . . . = 𝜆𝑘 = 0 1, 𝑥, 𝑥2, . . . , 𝑥𝑘−1
∙ the general solution of the homogeneous linear equation is:
𝑦(𝑥) = 𝐶1 · 𝑦1(𝑥) + 𝐶2 · 𝑦2(𝑥) + . . . + 𝐶𝑛 · 𝑦𝑛(𝑥),
where 𝑦1, 𝑦2, . . . , 𝑦𝑛 are the 𝑛 linear independednt solutions of the fundamentalsystem.
Nonhomogeneous linear equations with constant coefficients:
𝑦(𝑛) + 𝑎1𝑦(𝑛−1) + . . . + 𝑎𝑛−1𝑦
′ + 𝑎𝑛𝑦 = 𝑓(𝑥)
Algorithm:∙ find the general solution 𝑦(𝑥) of the attached homogeneous equation∙ the general solution for the nonhomogeneous equation will be:
𝑦(𝑥) = 𝑦(𝑥) + 𝑦𝑝(𝑥)
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where 𝑦𝑝(𝑥) is a particular solution of the nonhomogeneous equation which canbe found using the variation of constants method:�
if 𝑦(𝑥) = 𝐶1 ·𝑦1(𝑥)+𝐶2 ·𝑦2(𝑥)+ . . .+𝐶𝑛 ·𝑦𝑛(𝑥) is a general solution of theattached homogeneous quations we search for a particular solution of the form:
𝑦𝑝(𝑥) = 𝐶1(𝑥) · 𝑦1(𝑥) + 𝐶2(𝑥) · 𝑦2(𝑥) + . . . + 𝐶𝑛(𝑥) · 𝑦𝑛(𝑥)
�
one finds these functions 𝐶1(𝑥), 𝐶2(𝑥), . . . , 𝐶𝑛(𝑥) solving the linear system:⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩
𝐶′
1(𝑥) · 𝑦1 + 𝐶′
2(𝑥) · 𝑦2 + . . . + 𝐶′
𝑛(𝑥) · 𝑦𝑛 = 0
𝐶′
1(𝑥) · 𝑦′
1 + 𝐶′
2(𝑥) · 𝑦′
2 + . . . + 𝐶′
𝑛(𝑥) · 𝑦′
𝑛 = 0
𝐶′
1(𝑥) · 𝑦′′
1 + 𝐶′
2(𝑥) · 𝑦′′
2 + . . . + 𝐶′
𝑛(𝑥) · 𝑦′′
𝑛 = 0
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
𝐶′
1(𝑥) · 𝑦(𝑛−2)1 + 𝐶
′
2(𝑥) · 𝑦(𝑛−2)2 + . . . + 𝐶
′
𝑛(𝑥) · 𝑦(𝑛−2)𝑛 = 0
𝐶′
1(𝑥) · 𝑦(𝑛−1)1 + 𝐶
′
2(𝑥) · 𝑦(𝑛−1)2 + . . . + 𝐶
′
𝑛(𝑥) · 𝑦(𝑛−1)𝑛 = 𝑓(𝑥)
Particular cases:
∙ for some particular forms of 𝑓(𝑥) one can straightforward build a particularsolution 𝑦𝑝(𝑥) following the next rules:
1. if 𝑓(𝑥) = 𝑒𝛼𝑥𝑃𝑚(𝑥) where 𝑃𝑚 is a polynomial of degree 𝑚 and 𝛼 is nota root of the characteristic equation, then:
𝑦𝑝(𝑥) = 𝑒𝛼𝑥𝑄𝑚(𝑥)
for 𝑄𝑚 a 𝑚-th degree polynomial which has to be determined2. if 𝑓(𝑥) = 𝑒𝛼𝑥𝑃𝑚(𝑥) and 𝛼 is a multiple root of order 𝑘 for the charac-
teristic equation, we search for:
𝑦𝑝(𝑥) = 𝑥𝑘𝑒𝛼𝑥𝑄𝑚(𝑥)
3. if 𝑓(𝑥) = 𝑒𝛼𝑥 [𝑃𝑚(𝑥) cos𝛽𝑥 + 𝑄𝑚 sin𝛽𝑥] and 𝛼 + 𝑖𝛽 is not a root forthe characteristic equation, we search for:
𝑦𝑝(𝑥) = 𝑒𝛼𝑥 [𝑅𝑚(𝑥) cos𝛽𝑥 + 𝑆𝑚(𝑥) sin𝛽𝑥]
4. if 𝑓(𝑥) = 𝑒𝛼𝑥 [𝑃𝑚(𝑥) cos𝛽𝑥 + 𝑄𝑚 sin𝛽𝑥] and 𝛼 + 𝑖𝛽 is a multiple root oforder 𝑘 for the characteristic equation, we search for:
𝑦𝑝(𝑥) = 𝑥𝑘𝑒𝛼𝑥 [𝑅𝑚(𝑥) cos𝛽𝑥 + 𝑆𝑚(𝑥) sin𝛽𝑥]
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Solved problems
Problem 1. Find the general solutions for the next linear differentialequations:
a) 𝑦𝑖𝑣 − 5𝑦′′ + 4𝑦 = 0,
b) 𝑦′′′ − 6𝑦′′ + 12𝑦′ − 8𝑦 = 0,
c) 𝑦𝑖𝑣 + 5𝑦′′ + 4𝑦 = 0.
Solution: a) For the homogeneous equation 𝑦𝑖𝑣 −5𝑦′′ + 4𝑦 = 0 we write thecharacteristic equation:
𝑟4 − 5𝑟2 + 4 = 0𝑟2=𝑡⇒ 𝑡2 − 5𝑡 + 4 = 0
⇒ (𝑡− 1) (𝑡− 4) = 0
thus:
𝑟2 = 1, 𝑟2 = 4 ⇔𝑟1,2 = ±1, 𝑟3,4 = ±2.
Since the four roots are real and distinct we build the general solution of theaforementioned equation:
𝑦 (𝑥) = 𝑐1𝑒𝑥 + 𝑐2𝑒
−𝑥 + 𝑐3𝑒2𝑥 + 𝑐4𝑒
−2𝑥.
b) Let’s write the characteristic equation:
𝑟3 − 6𝑟2 + 12𝑟 − 8 = 0 ⇔ (𝑟 − 2)(︀𝑟2 − 4𝑟 + 4
)︀= 0
⇔ (𝑟 − 2)3
= 0.
We get a triple root which generates the following form for the general solution:
𝑦 (𝑥) = 𝑐1𝑒2𝑥 + 𝑐2𝑥𝑒
2𝑥 + 𝑐3𝑥2𝑒2𝑥
=(︀𝑐1 + 𝑐2𝑥 + 𝑐3𝑥
2)︀𝑒2𝑥.
c) For this equation the characteristic equation will be:
𝑟4 + 5𝑟2 + 4 = 0𝑟2=𝑡⇒ 𝑡2 + 5𝑡 + 4 = 0
⇒ (𝑡 + 1) (𝑡 + 4) = 0
hence:
𝑟2 = −1, 𝑟2 = −4 ⇔𝑟1,2 = ±𝑖, 𝑟3,4 = ±2𝑖.
The roots are complex and distinct and will generate the general solution:
𝑦 (𝑥) = 𝑐1cos𝑥 + 𝑐2sin𝑥 + 𝑐3cos 2𝑥 + 𝑐4sin 2𝑥.
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Problem 2. Find the general solutions for the following nonhomogeneouslinear equations:
a) 𝑦′′ − 𝑦′ − 2𝑦 = 3𝑒2𝑥,
b) 𝑦′′′ + 2𝑦′′ − 𝑦′ − 2𝑦 = 2 + 𝑒𝑥 + sin𝑥.
Solution: a) We write down the homogeneous equation attached to thisproblem 𝑦′′ − 𝑦′ − 2𝑦 = 0 which has the characteristic equation:
𝑟2 − 𝑟 − 2 = 0 ⇔(𝑟 − 2) (𝑟 + 1) = 0 ⇒ 𝑟1 = 2, 𝑟2 = −1.
The general solution of the homogeneous equation will be:
𝑦 (𝑥) = 𝑐1𝑒2𝑥 + 𝑐2𝑒
−𝑥.
Let us observe that the function 𝑓 (𝑥) = 3𝑒2𝑥 has 𝛼 = 2 which a a root for thecharacteristic equation (𝑟1 = 2) thus we have to search for a particular solutionof the form:
𝑦𝑝 (𝑥) = 𝑥1 · 𝑒2𝑥 · 𝑐.
We compute:
𝑦′𝑝 (𝑥) = 𝑐 (1 + 2𝑥) 𝑒2𝑥,
𝑦′′𝑝 (𝑥) = 4𝑐 (1 + 𝑥) 𝑒2𝑥.
and we substitute in the nonhomogeneous equation to obtain:
4𝑐 (1 + 𝑥) 𝑒2𝑥 − 𝑐 (1 + 2𝑥) 𝑒2𝑥 − 2𝑐𝑥𝑒2𝑥 = 3𝑒2𝑥 |: 𝑒2𝑥 ⇔
4𝑐 (1 + 𝑥) − 𝑐 (1 + 2𝑥) − 2𝑐𝑥 = 3
⇒ 𝑐 = 1 ⇒ 𝑦𝑝 (𝑥) = 𝑥𝑒2𝑥
Eventually, we can display the general solution of the given equation:
𝑦 (𝑥) = 𝑦 (𝑥) + 𝑦𝑝 (𝑥)
= 𝑐1𝑒2𝑥 + 𝑐2𝑒
−𝑥 + 𝑥𝑒2𝑥.
b) The homogeneous equation 𝑦′′′ + 2𝑦′′− 𝑦′− 2𝑦 = 0 has the characteristicequation
𝑟3 + 2𝑟2 − 𝑟 − 2 = 0 ⇔ (𝑟 − 1) (𝑟 + 1) (𝑟 + 2) = 0
⇒ 𝑟1 = 1, 𝑟2 = −1, 𝑟3 = −2,
with the general solution:
𝑦 (𝑥) = 𝑐1𝑒𝑥 + 𝑐2𝑒
−𝑥 + 𝑐3𝑒−2𝑥.
The term which is responsible for the nonhomogeneity 𝑓 (𝑥) = 2 + 𝑒𝑥 + sin𝑥has to be decomposed in:
2 + 𝑒𝑥 + sin𝑥
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in order to be able to use the particular classes studied in the beginning of thesenotes:
2 = 2 · 𝑒0·𝑥 ⇒ 𝛼 = 0
𝑒𝑥 = 𝑒1·𝑥 ⇒ 𝛼 = 1
sin𝑥 = 𝑒0·𝑥 sin (1 · 𝑥) ⇒ 𝛼 = 0 ± 1 · 𝑖.
One of these values (𝛼 = 1) is also a solution for the characteristic equation,thus the particular solution we should search for is:
𝑦𝑝 (𝑥) = 𝑎 + 𝑐𝑥𝑒𝑥 + 𝛼 cos𝑥 + 𝛽 sin𝑥.
We compute now:
𝑦′𝑝 (𝑥) = 𝑐𝑒𝑥 + 𝑐𝑥𝑒𝑥 − 𝛼 sin𝑥 + 𝛽 cos𝑥,
𝑦′′𝑝 (𝑥) = 2𝑐𝑒𝑥 + 𝑐𝑥𝑒𝑥 − 𝛼 cos𝑥− 𝛽 sin𝑥,
𝑦′′′𝑝 (𝑥) = 3𝑐𝑒𝑥 + 𝑐𝑥𝑒𝑥 + 𝛼 sin𝑥− 𝛽 cos𝑥.
and we substitute these results in the nonhomogeneous equation to obtain thecoefficients:
𝑎 = −1 𝑐 =1
6, 𝛽 = −1
5, 𝛼 =
1
10,
yielding the particular solution:
𝑦𝑝 (𝑥) = −1 +1
6𝑥𝑒𝑥 +
1
10cos𝑥− 1
5sin𝑥,
Finally, the general solution of the problem is:
𝑦 (𝑥) = 𝑦 (𝑥) + 𝑦𝑝 (𝑥) =
= 𝑐1𝑒𝑥 + 𝑐2𝑒
−𝑥 + 𝑐3𝑒−2𝑥 − 1 +
1
6𝑥𝑒𝑥 +
1
10cos𝑥− 1
5sin𝑥.
Problem 3. Solve the Cauchy problem:⎧⎨⎩ 𝑦𝑖𝑣 − 𝑦 = 𝑥3 + 𝑥
𝑦 (0) = 𝑦′ (0) = 𝑦′′ (0) = 𝑦′′′ (0) = 0
Solutieon: The characteristic equation is:
𝑟4 − 1 = 0 ⇔(︀𝑟2 − 1
)︀ (︀𝑟2 + 1
)︀= 0 ⇔ (𝑟 − 1) (𝑟 + 1) (𝑟 − 𝑖) (𝑟 + 𝑖) = 0
thus:𝑟1 = 1, 𝑟2 = −1, 𝑟3 = 𝑖, 𝑟4 = −𝑖
and will generate the solution:
𝑦 (𝑥) = 𝑐1𝑒𝑥 + 𝑐2𝑒
−𝑥 + 𝑐3cos𝑥 + 𝑐4sin𝑥.
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For the function 𝑓 (𝑥) = 𝑥3 + 𝑥 = 𝑒0·𝑥(︀𝑥3 + 𝑥
)︀of course 𝛼 = 0 is not a root
of the characteristic equation, hence:
𝑦𝑝 (𝑥) = 𝑎𝑥3 + 𝑏𝑥2 + 𝑐𝑥 + 𝑑
We start to compute:𝑦′𝑝 (𝑥) = 3𝑎𝑥2 + 2𝑏𝑥 + 𝑐,
𝑦′′𝑝 (𝑥) = 6𝑎𝑥 + 2𝑏, 𝑦′′′𝑝 (𝑥) = 6𝑎, 𝑦𝑖𝑣 (𝑥) = 0.
in order to substitute in the nonhomogeneous equations to obtain:
𝑎 = −1, 𝑏 = 0, 𝑐 = −1, 𝑑 = 0
thus the particular solution
𝑦𝑝 (𝑥) = −1 · 𝑥3 + 0 · 𝑥2 − 1 · 𝑥 + 0
= −𝑥3 − 𝑥.
Finally the formula for the general solution is:
𝑦 (𝑥) = 𝑦 (𝑥) + 𝑦𝑝 (𝑥) =
= 𝑐1𝑒𝑥 + 𝑐2𝑒
−𝑥 + 𝑐3 cos𝑥 + 𝑐4 sin𝑥−𝑥3 − 𝑥.
In order to solve the Cauchy problem we have to make use of the initial datato find the constants 𝑐1, 𝑐2, 𝑐3, 𝑐4 :
𝑦 (0) = 𝑐1 + 𝑐2 + 𝑐3,
𝑦′ (𝑥) = 𝑐1𝑒𝑥 − 𝑐2𝑒
−𝑥 − 𝑐3 sin𝑥 + 𝑐4 cos𝑥− 3𝑥2 − 1 ⇒ 𝑦′ (0) = 𝑐1 − 𝑐2 + 𝑐4 − 1,
𝑦′′ (𝑥) = 𝑐1𝑒𝑥 + 𝑐2𝑒
−𝑥 − 𝑐3 cos𝑥− 𝑐4 sin𝑥− 6𝑥 ⇒ 𝑦′′ (0) = 𝑐1 + 𝑐2 − 𝑐3,
𝑦′′′ (𝑥) = 𝑐1𝑒𝑥 − 𝑐2𝑒
−𝑥 + 𝑐3 sin𝑥− 𝑐4 cos𝑥− 6 ⇒ 𝑦′′′ (0) = 𝑐1 − 𝑐2 − 𝑐4 − 6.
Using:𝑦 (0) = 𝑦′ (0) = 𝑦′′ (0) = 𝑦′′′ (0) = 0
yields: ⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩
𝑐1 + 𝑐2 + 𝑐3 = 0
𝑐1 − 𝑐2 + 𝑐4 = 1
𝑐1 + 𝑐2 − 𝑐3 = 0
𝑐1 − 𝑐2 − 𝑐4 = 6
with the solution:
𝑐1 =7
4, 𝑐2 = −7
4, 𝑐3 = 0, 𝑐4 = −5
2.
Finally, the Cauchy problem has the solution:
𝑦 (𝑥) =7
4𝑒𝑥 − 7
4𝑒−𝑥 − 5
2sin𝑥−𝑥3 − 𝑥,
or equivalently:
𝑦 (𝑥) =7
2cosh𝑥− 5
2sin𝑥− 𝑥3 − 𝑥.
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Proposed problems
Problem 1. Find the general solutions for the following linear homogeneousdifferential equations with constant coefficients:
a) 64𝑦(8) + 48𝑦(6) + 12𝑦(4) + 𝑦(2) = 0,
b) 𝑦𝑖𝑣 − 3𝑦′′′ + 5𝑦′′ − 3𝑦′ + 4𝑦 = 0.
Problem 2. Find the general solutions for the following linear nonhomogeneousdifferential equations with constant coefficients:
a) 𝑦′′ − 4𝑦′ + 4𝑦 = 1 + 𝑒𝑥 + 𝑒2𝑥,
b) 𝑦′′ − 𝑦 = 𝑥𝑒𝑥 sin𝑥,
c) 𝑦𝑖𝑣 − 4𝑦′′ = 1,
d) 𝑦′′′ − 𝑦′′ = 𝑥.
f) 𝑦′′ − 3𝑦′ + 2𝑦 = 𝑥2𝑒𝑥
g) 𝑦′′ − 4𝑦 = 𝑒2𝑥(11 cos𝑥− 7 sin𝑥)
h) 𝑦′′ − 2𝑦′′ = 𝑒𝑥((4 − 4𝑥) cos𝑥− (6𝑥 + 2) sin𝑥)
i) 𝑦′′′ − 𝑦′′ + 𝑦′ − 𝑦 = cos𝑥
j) 𝑦′′ − 3𝑦′ + 2𝑦 = 𝑒3𝑥(𝑥2 + 𝑥)
k) 𝑦′′ − 5𝑦′ + 6𝑦 = 6𝑥2 − 10𝑥 + 2
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Bibliography
[1] Dennis. G. Zill. A First Course in Differential Equations with ModelingApplications, Brooks/Cole, 2013.
[2] Octavian Lipovan. Matematici speciale: Ecuatii diferentiale si teoria cam-purilor, Editura Politehnica, 2007.
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