Math 240 Homework Solution Set #1 Fall, 2009 Assigned ...

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Math 240 Homework Solution Set #1 Fall, 2009 Assigned ...

Math 240 Homework Solution Set #1 Fall, 2009Assigned September 9. Due September 16. The four numbered problems have equal credit.1a) Find two solutions, y 1 (t) and y 2 (t), of the differential equationy ′′ + y ′ − 6y = 0,such thaty 1 (0) = 1y 1(0) ′ = 0y 2 (0) = 0y 2(0) ′ = 1.1b) Same as 1a), but for the differential equationy ′′ + 10y ′ + 25y = 0.Solutions. a) The roots of x 2 + x − 6 are x = −3 and x = 2, so the “primitive” solutions ofthe D.E. are e −3t and e 2t . The sought-for solutions will have the form y 1 (t) = ae −3t + be 2tand y 2 (t) = ce −3t + de 2t , for some real numbers a, b, c, and d, such thaty 1 (0) = 1 = a + by ′ 1(0) = 0 = −3a + 2by 2 (0) = 0 = c + dy ′ 2(0) = 1 = −3c + 2d.Elementary linear algebra implies a = 2/5, b = 3/5, c = −1/5, d = 1/5. b) Now the“primitive” solutions are e −5t and te −5t . We seek a, b, c, and d such that y 1 (t) = ae −5t +bte −5t and y 2 (t) = ce −5t + dte −5t satisfy the initial conditions; i.e.:y 1 (0) = 1 = ay ′ 1(0) = 0 = −5a + by 2 (0) = 0 = cThe solutions are a = 1, b = 5, c = 0, d = 1.y ′ 2(0) = 1 = −5c + d.2. Let a, b, and c be real numbers, with a and b not both equal to 0, and set f(t) ≡a cos(ct) + b sin(ct). Show thatf(t) = A cos(ct − φ),


whereA = (a 2 + b 2 ) 1/2cos φ = a Asin φ = b A .Solution. The cosine subtraction formula says cos(α−β) = cos α cos β +sin α sin β. ThereforeA cos(ct − φ) = A(cos(ct) cos φ + sin(ct) sin φ), which equals a cos(ct) + b sin(ct) whenA and φ are as given. Note that our definition of A implies that an appropriate φ doesexist!3a) Find two solutions, y 1 (t) and y 2 (t), of the differential equationt 2 y ′′ + ty ′ − 4y = 0,(valid for t > 0) such thaty 1 (1) = 1y 1(1) ′ = 0y 2 (1) = 0y 2(1) ′ = 1.3b) Same as 3a), but for the differential equationt 2 y ′′ + 3ty ′ + y = 0.Solutions. a) The roots of x(x − 1) + x − 4 are x = ±2; therefore the “primitive” solutionsof the D.E. are t 2 and t −2 . We set y 1 (t) = at 2 + bt −2 and y 2 (t) = ct 2 + dt −2 , and seek todetermine the coefficients so thaty 1 (1) = 1 = a + by ′ 1(1) = 0 = 2a − 2by 2 (1) = 0 = c + dy ′ 2(1) = 1 = 2c − 2d.The solutions are a = b = 1/2, c = 1/4, d = −1/4. b) Now the “primitive” solutions aret −1 and t −1 log t. We set y 1 (t) = at −1 + bt −1 log t and y 2 (t) = ct −1 + dt −1 log t, and pickthe coefficients so thaty 1 (1) = 1 = ay ′ 1(1) = 0 = −a + by 2 (1) = 0 = cy ′ 2(1) = 1 = −c + d.


The solutions are a = b = 1, c = 0, d = 1.4. Solve the differential equation,t 2 y ′′ + ty ′ + 4y = 0,(valid for t > 0) subject to the initial condition y(1) = y ′ (1) = 1.Solution. The “primitive” solutions are t 2i and t −2i , but what do these mean? If a ∈ Rthen t ia = e ia log t = cos(a log t) + i sin(a log t). Therefore t 2i and t −2i are linear combinationsof cos(2 log t) and sin(2 log t), and we can write our solution y(t) = a cos(2 log t) +b sin(2 log t). We choose a and b so thatThus, a = 1 and b = 1/2.y(1) = 1 = ay ′ (1) = 1 = 2b.

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