Download Math 2214 Test 2 Sample Problems: Solving Differential Equations - Prof. Evgeny Savel'Ev and more Exams Differential Equations in PDF only on Docsity!
Math 2214 Test 2 Sample Problems
- Find a general solution to the following equations:
(a) y′′^ − 2 y′^ + 5y = 0 Characteristic equation is λ^2 − 2 λ + 5 = 0. You can write it as (λ − 1)^2 + 2^2 = 0 The roots are λ 1 = 1 + 2i and λ 2 = 1 − 2 i Thus the general solution is: yG(t) = c 1 et^ cos(2t) + c 2 et^ sin(2t) (b) y′′^ − 2 y′^ + y = 0 Characteristic equation is λ^2 − 2 λ + 1 = 0. You can write it as (λ − 1)^2 = 0 The roots are repeated: λ 1 , 2 = 1 Thus the general solution is: yG(t) = c 1 et^ + c 2 tet (c) y′′^ − 2 y = 0 Characteristic equation is λ^2 − 2 = 0. The roots are: λ 1 , 2 = ±
Thus the general solution is: yG(t) = c 1 et
√ 2
√ 2
(d) y′′^ − 4 y′^ + 20y = 0 Characteristic equation is λ^2 − 4 λ + 20 = 0. The roots are λ 1 = 2 + 4i and λ 2 = 2 − 4 i Thus the general solution is: yG(t) = c 1 e^2 t^ cos(4t) + c 2 e^2 t^ sin(4t)
- For the non-homogeneous equation y′′^ − 4 y′^ + 20y = g(t), indicate the correct guess for the particular solution yp for each of the following forms of g(t).
(a) g(t) = 4t + cos(4t) yp(t) = A 1 t + B 1 + A 2 cos(4t) + A 3 sin(4t) (b) g(t) = 12e−^2 t^ sin(4t) yp(t) = A 1 e−^2 t^ sin(4t) + A 2 e−^2 t^ cos(4t) (c) g(t) = t^2 + t + 1 yp(t) = At^2 + Bt + C (d) g(t) = te^2 t^ cos(4t) yp(t) = t(A 1 t + B 1 )e^2 t^ cos(4t) + t(A 2 t + B 2 )e^2 t^ sin(4t)
- The characteristic equation for a certain 7th order differential equation has the follwong factorization:
(λ − 1)^2 (λ^2 − 1)(λ^3 + 8) = 0
Write down the general solution to this equation. (λ − 1)^2 gives us the root λ = 1 twice. (λ^2 − 1) gives us other two roots λ = ±1. For (λ^3 + 8) you either factor it out (λ^3 + 8) = (λ + 2)(λ^2 − 2 λ + 4) or use this:
λ^3 + 8 = 0 ⇒ λ = (−8)^1 /^3 ⇒ λ = (2eiπ+i^2 πk)^1 /^3 ⇒ λ = 2eiπ/^3 λ = 2ei^3 π/^3 λ = 2ei^5 π/^3
Applying Euler’s formula and writing all the roots in a row we get:
λ ∈ { 1 , 1 , 1 , − 1 , 1 + i
3 , − 2 , 1 − i
Thus the general solution is: yG(t) = c 1 et^ + c 2 tet^ + c 3 t^2 et^ + c 4 e−t^ + c 5 e−^2 t^ + c 6 et^ cos(t
- Find the general solution to the non-homogeneous equation y′′^ − 2 y′^ + 5y = 3 cos(2t) + 2e−t^ and solve the initial value problem if y(0) = 0 and y′(0) = 0. Method of undetermined coefficients is easier to apply here. For the term 3 cos(2t) we should introduce tr(A cos(2t) + B sin(2t)) into the guess for a particular solution. Since cos(2t) doesn’t appear in the complementary solution (see Prob. 1.a), r = 0 and therefore tr^ = 1. For the term 2e−t^ we include the term Ce−t^ into yP. Thus:
yP =A cos(2t) + B sin(2t) + Ce−t^5 y = 5A cos(2t) + 5B sin(2t) + 5Ce−t y P′ =− 2 A sin(2t) + 2B cos(2t) − Ce−t^ ⇒ − 2 y′^ = 4A sin(2t) − 4 B cos(2t) + 2Ce−t y P′′ =− 4 A cos(2t) − 4 B sin(2t) + Ce−t^ y′′^ = − 4 A cos(2t) − 4 B sin(2t) + Ce−t
Plugging it into the equation and simplifying we get
(A − 4 B) cos(2t) + (B + 4A) sin(2t) + 8Ce−t^ = 3 cos(2t) + 2e−t
To find the coefficients we need to solve the system of equations
(A − 4 B) =3 A = 173 (4A + B) =0 ⇒ B = −^1217 8 C =2 C = (^14)
Thus the general solution is
yG(t) = c 1 et^ cos(2t) + c 2 et^ sin(2t) +
cos(2t) −
sin(2t) +
e−t
Last bit is to find the constants c 1 and c 2 to satisfy y(0) = y′(0) = 0. After some algebra you get: y(t) = −
17 +^
1 4
et^ cos(2t) +
34 +^
1 4
et^ sin(2t) + 173 cos(2t) − 1217 sin(2t) + 14 e−t
- Consider a linear differential equation
t^2 y′′^ − 3 ty′^ + 4y = 0 t > 0
(a) At least one of the solutions has the form y(t) = tk. Plug this into the equation and find all solutions of this form. Whether or not you divide by t^2 at the beginning wouldn’t make any difference here (for part (c) it will though), so I’ll proceed with the original equation.
y = tk y′^ = ktk−^1 y′′^ = k(k − 1)tk−^2
⇒ t^2 k(k − 1)tk−^2 − 3 t · ktk−^1 + 4tk^ = 0 ⇒ tk(k^2 − 4 k + 4) = 0
Since tk^ can never be zero (see the assumptions) we should have k^2 − 4 k + 4 = 0. This equation has two repeated roots k = 2. Therefore, the function y(t) = t^2 must be a solution (which is true, just check it). (b) In the previous part you should find that there is only one solution y 1 (t) = t^2. Use the method of reduction of order to find the second linearly independent solution. We assume that the second solution has the form y 2 (t) = u(t)·y 1 (t), where u(t) is a function that is yet to be determined. Write the expression for y 2 (t) and compute its first and second derivatives:
- Consider the linear differential equation
ty′′^ − (2t + 1)y′^ + (t + 1)y = 0 t > 0
(a) Show that the function y 1 (t) = et^ is a solution. To plug y 1 in we need its first and second derivatives: y 1 (t) = et^ ⇒ y′ 1 (t) = et^ ⇒ y 1 ′′ (t) = et Now: ty 1 ′′ − (2t + 1)y′ 1 + (t + 1)y 1 = 0 ⇒ tet^ − (2t + 1)et^ + (t + 1)et^ ≡ 0 After simplification we will see that the equality indeed holds (b) Use the method of reduction of order to find a second linearly independent solution. We assume that the second solution has the form y 2 (t) = u(t) · et, where u(t) is a function that is yet to be determined. Computing first and second derivatives gives: y 2 (t) = u(t) · et^ ⇒ y 2 ′(t) = u′(t) · et^ + u(t) · et^ ⇒ y 1 ′′ (t) = u′′(t) · et^ + 2u′(t) · et^ + u(t) · et
Plug y 2 (t) into the differential equation above and simplify the result: ty 2 ′′ − (2t + 1)y′ 2 + (t + 1)y 2 = 0 ⇒ t(u′′^ · et^ + 2u′^ · et^ + u · et)− − (2t + 1)(u′^ · et^ + u · et) + (t + 1)u · et^ ≡ 0 After simplification it becomes tetu′′^ − etu′^ ≡ 0 This can be simplified even further by dividing by tet: u′′^ −
t
u′^ = 0
Use the change of variables v(t) = u′(t) and v′(t) = u′′(t) to convert your equation into a first order equation. Solve the equation for v(t) and then find u(t) by integrating v(t): After change of variable we obtain: v′^ −
t
v = 0 We can solve it as a first order equation (solution has the form e−P^ (t), where P (t) =
−^1 t dt): v(t) = e−(−^ ln^ t)^ = t Therefore u(t) =
v(t) dt =
t^2 2
and finally y 2 (t) = u(t)et^ =
t^2 2
et
The Wronskian has the form: W (t) =
et^ t
2 2 e
t
et^ t
2 2 e
t (^) + tet
= te^2 t
The Wronskian is nonzero if t > 0, therefore y 1 and y 2 make up a fundamental set. Thus the general solution has the form: y(t) = c 1 et^ + c 2 t
2 2 e
t (^) or equivalently y(t) = c 1 et (^) + c 2 t (^2) et
- Solve the initial value problem y′′^ − 4 y′^ + 4y = 0, y(0) = 1, y′(0) = 2 Characteristic equation is λ^2 − 4 λ + 4 = 0. The root λ = 2 is repeated. Thus the general sooution is:
yG(t) = c 1 e^2 t^ + c 2 te^2 t
Solving for constants using initial condtions we get:
1 ≡ y(0) = c 1 + c 2 · 0 2 ≡ y′(0) = 2c 1 + c 2
c 1 = 1 c 2 = 0
Therefore, the solution to the initial value problem is y(t) = e^2 t
- Solve the initial value problem
y′′^ − 4 y′^ + 4y =
e^2 t t + 1
, y(0) = 1, y′(0) = 2
The complementary solution is yc = c 1 e^2 t^ + c 2 te^2 t The Wronskian is: W (t) = e^4 t
Equation has constant coefficients, but the right-hand side has a rational fraction in it, thus we must use the method of variation of parameters
Using the formulas for u′ 1 and u′ 2 we obtain:
u′ 1 =
−y 2 g(t) W
−te^2 t^ · e^2 t (t + 1)e^4 t^
−t t + 1
⇒ integrate ⇒ u 1 (t) = ln(t + 1) − t
u′ 2 =
y 1 g(t) W
e^2 t^ · e^2 t (t + 1)e^4 t^
t + 1
⇒ ⇒ u 2 (t) = ln(t + 1)
Therefore the particular solution has the form: yp(t) = u 1 · y 1 + u 2 · y 2 = (ln(t + 1) − t)e^2 t^ + t ln(t + 1)e^2 t
General solution is obtained by summing complementary and particular solutions: y(t) = c 1 e^2 t^ + c 2 te^2 t^ + (ln(t + 1) − t)e^2 t^ + t ln(t + 1)e^2 t^ = e^2 t
c 1 + (c 2 − 1)t + (t + 1) ln(t + 1)
We shall need a derivative for calculating unknown constants c 1 and c 2 : y′(t) = 2c 1 e^2 t^ + c 2 (e^2 t^ + 2tet)+
2 ln(t + 1) − 2 t +
t + 1
e^2 t^ +
ln(t + 1) +
t t + 1
e^2 t
The last task is to compute c 1 and c 2 using initial conditions: 1 ≡ y(0) = c 1 + (c 2 − 1) · 0 + 0 2 ≡ y′(0) = 2c 1 + c 2 + 0 + 0
c 1 = 1 c 2 = 0 Remark: It is just a concidence that the constants are the same as in problem 7. In general they should be different.
Final solution is: y(t) = e^2 t
1 − t + (t + 1) ln(t + 1)
I also recommend to revisit homework problems assigned for sections in chapter 3.
