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Instructor: Dr. Yuli B. Rudyak
MAP 2302, Differential Equations, Section 022G Spring 2017
QUIZ 2, 1 MARCH 2017
Problem 1. Solve the diffrential equation
(a) y 00 + 7y 0 + 6y = 0.
Answer: y = C1 e−x + C2 e−6x .
(b) y 00 + 4y 0 + 4y = 0.
Answer: y = (C1 + C2 x)e−2x .
Problem 2. Solve the initial value problem y 00 − 2y 0 + 2y = 0, y(0) = 0, y 0 (0) = 1.
Solution: We see that r1,2 = 1 ± i are the roots of the characteristic equation. So,
the general solution is y = ex (C1 cos x + C2 sin x). Now, y(0) = 0, and so C1 = 0.
Now, consider the solution y = C2 ex sin x with y 0 = C2 (ex sin x + ex cos x) and let
y 0 (0) = 1, we get C2 = 1. Thus, y = ex sin x.
Problem 3. Determine the form of a particular solution to the following differential
equations (do not evaluate coefficients).
(a) y 00 − 4y 0 = x + 1 + xe2x + e4x + e4x sin 4x.
Answer: y = C1 + C2 e4x + (Ax2 + Bx) + (Cx + D)e2x + Exe4x +
e4x (F sin 4x + G cos 4x)
(b) y 00 − 4y 0 + 4y = 2xe2x + x cos x + e2x sin x.
Answer: y = e2x (C1 x + C2 ) + x2 e2x (Ax + B) + (Cx + D) cos x + (Ex + F ) sin x +
e2x (G cos x + H sin x).
Problem 4. Solve the initial value problem y 00 −3y 0 +2y = ex ,
y(0) = 0, y 0 (0) = 0.
Solution: We have r1 = 1, r2 = 2 and hence yhom = C1 ex + C2 e2x . So, we look
for a particular solution yp = Axex and find A = −1. So, y = C1 ex + C2 e2x − xex .
Put y(0) = 0 and get C1 + C2 = 0. Put y 0 (0) = 0 and get C1 + 2C2 − 1 = 0. So,
C1 = −1, C2 = 1. Thus, y = e2x − ex − xex .
Problem 5. Find the general solution to the equation y 00 + y = sec x.
Solution: We use variation of parameters. So, yhom = C1 cos x + C2 sin x. We look
for y = v1 cos x + v2 sin x. Now, we have the system
v10 cos x + v20 sin x = 0,
−v10 sin x + v20 cos x = sec x.
We solve the system and get v10 = − sec x sin x, v20 = 1. So,
Z
Z
v1 = − (sec x sin x)dx = − tan xdx = C1 + ln | cos x|.
Futhermore, v2 = x + C2 . Thus, y = (C1 + ln | cos x|) cos x + (x + C2 ) sin x.
Problem 6. Find a general solution to the differential equation
x2 y 00 + xy 0 − y = 0, x > 0.
Solution: We put y = xr and get xr [r(r − 1) + r − 1] = 0. So, (r2 − 1) = 0, and so
r1 = 1 and r2 = −1. So, y1 = x, y2 = x−1 , and thus y = C1 x + C2 x−1 .
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