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570 CHAPTER 29. FIRST ORDER SCALAR ODE

(a) 1

(b) 2

(c) 3

(d) 4

(e) −2

(f) 12

(g) −3

(h) −2

3. Solve the following initial value problem. ty′− y = 1t2 ,y(1) = 2. Would it make any

sense to give the initial condition at t = 0?

4. Solve the following initial value problem. ty′+ y = 1t ,y(−1) = 2. Would it make

any sense to give the initial condition at t = 0? Hint: You need to remember that∫ 1t dt = ln |t|+C.

5. Solve the following initial value problems.

(a) ln(t)y′+ 1t y = ln(t) , y(2) = 3.

(b) ln(t)y′− 1t y = ln2 (t) , y(2) = 3.

(c) y′+ tan(t)y = cos3 (t) , y(0) = 4.

(d) cosh(t)y′+ sinh(t)y = sinh(t) , y(0) =−4

6. You have the equation y′+ p(t)y = q(t) where P′ (t) = p(t) . Give a formula for allsolutions to this differential equation.

7. The height of an object at time t is y(t) . It falls from an airplane at 30,000 feet whichis traveling East at 500 miles per hour and is acted on by gravity which we willassume has acceleration equal to 32 feet per second squared and air resistance whichwe will suppose yields an acceleration equal to .1 times the speed of the falling objectopposite to the direction of motion. If its initial velocity is in the direction of motionof the airplane, find a formula for the position of the object as a function of t in feet.

8. Solve the following differential equations. Give the general solution.

(a)(x3 + y

)dx− xdy = 0

(b) ydx+(x− y)dy = 0 Hint: You might look for x as a function of y.

(c) y(y2− x

)dy = dx

(d) 2ydx =(x2−1

)(dx−dy)

(e) L didt +Ri = E sin(ωt) . Here L,R,E are positive constants. L symbolizes induc-

tance and R resistance while i is the current.

9. For compounding interest n times in one year which has interest rate r per year, theamount after t years is given by A0

(1+ r

n

)tn. Show that

limn→∞

(1+

rn

)tn= ert ,

thus giving the same conclusion as mentioned in the chapter.

10. Consider the equation y′+2ty = t,y(0) = 32.76. Find limt→∞ y(t).

57010.CHAPTER 29. FIRST ORDER SCALAR ODE(a) 1 (e) —2(b) 2 (f) 12(c) 3 (g) —3(d) 4 (h) —2. Solve the following initial value problem. ty’ — y = wy (1) = 2. Would it make anysense to give the initial condition at t = 0?Solve the following initial value problem. ty’ + y = i, y(—1) = 2. Would it makeany sense to give the initial condition at t = 0? Hint: You need to remember thatJ tdt =In|t|+C.Solve the following initial value problems.(a) In(t)y’ + Fy =In(t), y(2) =3.(b) In(t)y'— Ly =n? (0), y(2) =3.(c) y’+tan(t)y = cos? (t), y(0) =4.(d) cosh (t) y’ +sinh(t)y = sinh(t), y(0) = —4You have the equation y’ + p(t) y = q(t) where P’ (t) = p(t). Give a formula for allsolutions to this differential equation.The height of an object at time f is y(t) . It falls from an airplane at 30,000 feet whichis traveling East at 500 miles per hour and is acted on by gravity which we willassume has acceleration equal to 32 feet per second squared and air resistance whichwe will suppose yields an acceleration equal to .1 times the speed of the falling objectopposite to the direction of motion. If its initial velocity is in the direction of motionof the airplane, find a formula for the position of the object as a function of f in feet.. Solve the following differential equations. Give the general solution.(a) (x8 +y) dx—xdy =0(b) ydx+(x—y)dy = 0 Hint: You might look for x as a function of y.(c) y(y? —x) dy = dx(d) 2ydx = (x — 1) (dx —dy)(e) Le +Ri=Esin(ot). Here L,R,E are positive constants. L symbolizes induc-tance and R resistance while 7 is the current.For compounding interest n times in one year which has interest rate r per year, theamount after ¢ years is given by Ao (1 + rym . Show thatr\tnlim (1+-) =e",n—oo nthus giving the same conclusion as mentioned in the chapter.Consider the equation y’ + 2ty =t,y (0) = 32.76. Find lim,_,.. y(t).