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351 AMATH Homework 2 Keys
July 8, 2009
Section2.1 14,16,19,31,40 Section2.4 3,12,15,29 Section2.6 8,21,23,24
Section 2.1
In each of Problems 13 through 20 nd the solution of the given initial value problem. 14. y + 2y = te2t , y (1) = 0 Multiply (t) = e2t , then
e2t y + 2e2t y ey e2t y y
2t
=t =t = t2 +C 2 t2 +C = e2t 2
Apply initial condition y (1) = So the solution is
1 2
1 + C e2 = 0 = C = 2 .
y = e2t
t2 1 2 2
16. y + (2/t)y = (cos t)/t2, y ( ) = 0, t > 0 2 Integrating factor (t) = e t dt = t2 . Multiply it to the DE, we get
y t2 + 2ty ty t y y
2 2
= = = =
cos t cos t sin t + C t2 (sin t + C )
1
Apply initial condition, we have
y ( ) = 2 (sin + C ) = C 2 = 0 = C = 0
Plug C back into the general solution,
y (t) = t2 sin t
19. t3 y + 4t2 y = et , y (1) = 0, t < 0 First write it in the standard form
4 y + y = t3 et t 4 (t) = e t dt = t4
then multiply it,
t4 y + 4t3 y t4 y t4 y y = = = = tet te
t
tet dt = tet et + C e4 tet et + C
Apply initial condition
y (1) = e4 (e e + C ) = e4 C = 0 = C = 0
Plug it into the general solution, we get
y (t) = e4t (t + 1)
31. Consider the initial value problem
3 y y = 3t + 2et , 2 y (0) = y0 .
Find the value of y0 that separates solutions that grow positively as t from those that grow negatively. How does the solution that corresponds to this critical value of y0 behave as t ? Integrating factor 3 3
(t) = e
2 dt
= e 2 t
2
So
e 2 t y e 2 t y
3 3
3t + 2et e 2 t = 3te 2 t + 2e 2 3 t 2t = 3te dt + 2 e 2 dt 3 3 t = 2 te 2 t e 2 t dt 4e 2 =
3
3
t
3 t 43 = 2te 2 t e 2 t 4e 2 + C 3 3 4 y (t) = 2t 4et + Ce 2 t 3
4 Apply initial condition y (0) = 3 4 C = y0 = C = y0 + 16 . So the specic 3 solution is
y (t) = 2t
4 16 3 4et + (y0 + )e 2 t 3 3
where the last term is the dominant term. Thus the critical value of y0 is the value that makes it 0, i.e.
y0 + 16 16 = 0 = y0 = 3 3
when y0 > 16 , y (t) as t + 3 when y0 < 16 , y (t) + as t + 3 when y0 = 16 , y (t) = 2t 3
4 3
4et as t +
In Problem 40 use the method of Problem 38 to solve the given dierential equation. 40. y + (1/t)y = 3 cos 2t, t>0 Based on 38, we want to nd out A(t) that satises
A (t) A(t) 3 cos 2t e t dt = 3t cos 2t 3 3 = 3t cos 2tdt = t sin 2t + cos 2t + C 2 4 =
1
So the general solution is
y (t) = A(t)e =
1 t dt
=
A(t) t
3 3 C sin 2t + cos 2t + 2 4t t
3
Section 2.4
In each of Problems 1 through 6 determine (without solving the problem) and interval in which the solution of the given initial value problem is certain to exist. 3. y + (tan t)y = sin t, y ( ) = 0 This is a rst order linear DE. p(t) = t tan t is continuous on k , k + 2 2 where k is an integer. q (t) = sin t is continuous on (, +). According to Thm 2.4.1, the solution to this IVP exists on , 32 . 2 In each of Problems 7 through 12 state where in the ty -plane the hypotheses of Teorem 2.4.2 are satised. 12. dy = (cot t)y dt 1+y Function f (t, y ) is continuous in (k, k + ) (, 1) and (k, k + ) (1, +). f cot t y = (1+y )2 , which is continuous in the same area as that of f (t, y ). So Thm 2.4.2 is satised when
t = k, k Z and y = 1
In each of Problems 13 through 16 solve the given initial value problem determine and how the interval in which the solution exists depends on the initial value y0 . 15. y + y 3 = 0, y (0) = y0
y = y 3
This is a separable equation, we need to divide y 3 , then let's check the case of y (x) 0. Obviously, it is a solution. If y (t) = 0,
2 dy y3 y 2 y
2
= = =
2dx 2x + C 1 2x + C
For the initial condition,
If y0 > 0,
2 y0 =
1 1 C= 2 y= C y0
y0
2 2xy0 + 1
If y0 < 0,
2 y0 =
1 1 C= 2 y= C y0
1 2x +
1 2 y0
=
2 y0 2+1 = 2xy0
y0
2 2xy0 + 1
4
If y0 = 0, then y falls into the rst case we talked above: y (x) 0.
Now let's analyze the interval on which the solution exists,
1 1 2 If y0 = 0, 2xy0 + 1 > 0 x > 2y2 2y2 , + ;
0 0
If y0 = 0, no ristriction on x (, +).
Sometimes it is possible to solve a nonlinear equation by making a change of the dependent variable that converts it into a linear equation. The most important such equation has the form
Bernoulli Equations.
y + p(t)y = q (t)y n .
and is called a Bernoulli equation after Jakob Bernoulli. Problem 27 through 31 deal with equations of this type. 29. y = ry ky 2 , r > 0 and k > 0. 1 Substitute v = y 1 , then y = v , dy = v12 dv . Now the DE becomes dt dt
1 dv 1 r v 2 dt v dv + rv dt = k =k 1 v2
Integrating factor (t) = ert , multiply it to the above equation,
ert v v (t) y (t) = = = kert k + Cert r 1 r = v (t) k + Crert
Section 2.6
Determin whether each of the equations in Problems 1 through 12 is exact. If it is exact, nd the solution. 8. (ex sin y + 3y )dx (3x ex sin y )dy = 0 Check
P y
Q
Q x
=
ex cos y + 3 (3 + ex sin y ) ex cos y + 6 ex sin y = 3x + ex sin y 3x + ex sin y
Q x
Check
P
P y
=
ex cos y 6 + ex sin y ex sin y + 3y
So it's not an exact equation. 5
Show that the equations in Problems 19 through 22 are not exact but become exact when multiplied by the given integrating factor. Then solve the equations. 21. ydx + (2x yey )dy = 0, (x, y ) = y Check
Q P =1= =2 y x
which is not exact. Multiply (x, y ) = y , the DE becomes
y 2 dx + 2xy y 2 ey dy = 0
Check which means exact now.So
P Q = 2y = = 2y y x
x y
= y2 = 2xy y 2 ey
From the rst equation, we have (x, y ) = xy 2 + (y ) , then
= 2xy + (y ) = 2xy y 2 ey y = (y ) = y 2 ey + 2yey 2ey
So the nal solution is
(x, y ) = xy 2 y 2 ey + 2yey 2ey = C
23. Show that if (Nx My )/M = Q, where Q is a function of y only, then the dierential equation
M + Ny = 0
has an integrating factor of the form
(y ) = exp
Q(y )dy.
If the integrating factor is (y ), we multiply it to the equation and get,
M dx + N dy = 0
6
which now should be an exact equation, i.e.
(M ) y d M M + dy y 1 d dy = = = (N ) x N x Nx M y =Q M Q(y )dy exp Q(y )dy
ln || = (y ) =
24. Show that if (Nx My )/(xM yN ) = R, where R depends on the quantity xy only, then the dierential equation
M + Ny = 0
has an integrating factor of the form (xy ). Find a general formula for this integrating factor. If the integrating factor is denoted by (xy ), multiply it to the DE we should get an exact equation,
M dx + N dy = 0
then
(M ) y M M + y x = = (N ) x N N + x y N M x y N y (xy ) M x (xy ) (yN xM ) M x Ny =R yN xM xy exp R (s) ds
(Mx Ny ) = = = =
=
7

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