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WSU Math 315 Sample Problems for Test I (Fall, 2007)
Solutions
Note: The ODE in Problem 13 has been changed from that given out in class.
- For the differential equation y′^ = y(y − 2)
(a) Is the equation linear? (b) Is the function y = 2 a solution? (c) What is the slope of a solution at (1, 3)? Solution (a) First order linear equations have the form:
y′^ + p(t)y = g(t),
where p(t), g(t) are given functions. Here, we clearly do not have this form, since upon simplifying we obtain: y′^ + 2y = y^2 , and the right-hand side is a function of y rather than of t. So, the equation is not linear, i.e. nonlinear. In general, any equation containing terms like y^2 , √y, (^1) y , sin(y), etc., cannot be written in the linear form y′^ + p(t)y = g(t), and so such an equation is nonlinear. (b) Considering y = 2, y′^ = 0, hence using the given ODE, we have 0 = 2(2 − 2) = 0; hence y = 2 satisfies the equation, and thus is a solution. (c) At the point (x 1 , y 1 ) = (1, 3), observe that substituting y 1 = 3 into the equation yields: y′^ = 3(3 − 2) = 3. So, the slope of a solution at (1, 3) is 3.
- Solve the following first-order differential equation: dy dt
y cos(3t), y(0) = 1.
Solution. The equation is separable. Simplifying yields:
− cos(3t) + y−^1 /^2 y′^ = 0.
Taking M (t) = − cos(3t), N (y) = y−^1 /^2 , and recalling that the general solution of a separable equation M (t) + N (y)y′^ = 0 is implicitly defined by ∫ M (t) dt +
N (y) dy = c
where c is an arbitrary constant, here we obtain: ∫ M (t) dt +
N (y) dy = −
cos(3t) dt +
y−^1 /^2 dy
= −
sin(3t) + 2y^1 /^2
and thus an implicitly defined solution is:
−
sin(3t) + 2y^1 /^2 = c.
Using the initial condition y(0) = 1 yields c = 2, hence our solution becomes:
−
sin(3t) + 2y^1 /^2 = 2.
Notice further that we can solve the above equation for y to obtain:
y =
sin(3t)
sin(3t)
- Find the general solution of the following differential equation:
dy dt
= 3y + 5t.
Solution. The above equation is linear, as simplification reveals:
y′^ − 3 y = 5t
where p(t) = −3 and g(t) = 5t. We proceed as follows:
- Integrating factor: μ(t) = e
∫ (^) p(t) dt = e
∫ (^) − 3 dt = e−^3 t.
μ(t)g(t) dt =
5 te−^3 t^ dt = −
te−^3 t^ +
e−^3 t^ dt = −
e−^3 t
t +
- General solution: y = e^3 t
[
e−^3 t
t +
]
t +
- Consider the differential equation:
(xy^2 − sin(x))dx + x^2 ydy = 0.
(a) Verify the equation is exact. (b) Find the general solution. Solution. Notice that this equation has the form M (x, y)dx + N (x, y)dy = 0 where M (x, y) = xy^2 − sin(x), N (x, y) = x^2 y. (a) For this equation, we have My(x, y) = 2xy = Nx(x, y), so the equation is exact.
(b) Picard’s iteration method is the construction of a sequence of functions associ- ated with the above integral equation, each function of which satisfies the initial condition φ(0) = 0. Namely, letting φ 0 (t) = 0, the nth^ term of the sequence is:
φn(t) =
∫ (^) t
0
f (s, φn− 1 (s)) ds.
So, for f (t, y) = t^2 + y^2 , we have:
φ 1 (t) =
∫ (^) t
0
(s^2 + (φ 0 (s))^2 ) ds =
∫ (^) t
0
s^2 ds = t^3 3
φ 2 (t) =
∫ (^) t
0
(s^2 + (φ 1 (s))^2 ) ds =
∫ (^) t
0
s^2 + t^6 9
ds = t^3 3
t^7 63
- Solve the following initial-value problem:
y′′^ + 5y′^ + 6y = 0, y(0) = 1, y′(0) = − 1. Solution. The characteristic equation for the above 2nd order linear ODE is r^2 + 5r + 6 = (r + 3)(r + 2) = 0, which has the real, distinct roots r 1 = −3 and r 2 = −2. Thus the general solution is y(t) = c 1 er^1 t^ + c 2 er^2 t^ = c 1 e−^3 t^ + c 2 e−^2 t. Applying the initial condition y(0) = 1 gives the equation c 1 + c 2 = 1 and the initial condition y′(0) = −1 gives − 3 c 1 − 2 c 2 = − 1. From the first equation c 1 = 1 − c 2 ; substituting this into the second equation gives: −3(1 − c 2 ) − 2 c 2 = − 1 =⇒ −3 + c 2 = − 1 =⇒ c 2 = 2 =⇒ c 1 = − 1. So the solution to the above initial-value problem is: y(t) = −e−^3 t^ + 2e−^2 t.
- Find the fundamental set of solutions for the following problem, and find the Wronskian of the fundamental solution set at t = 0: y′′^ + 2y′^ + 2y = 0. Solution. The characteristic equation is r^2 + 2r + 2 = 0, whose solutions we obtain from the Quadratic Formula:
r =
− 2 ± 2 i 2 = − 1 ± 2 i.
So the characteristic equation has complex conjugate room − 1 ± 2 i, so putting λ = − 1 and μ = 2, the general solution of the above ODE is:
y(t) = c 1 eλt^ sin(μt) + c 2 eλt^ cos(μt) = c 1 e−t^ sin(2t) + c 2 e−t^ cos(2t).
A fundamental set of solutions is {y 1 , y 2 } = {e−t^ sin(2t), e−t^ cos(2t)}, whose Wronskian is:
W (y 1 , y 2 )(t) =
∣∣ e −t (^) sin(2t) e−t (^) cos(2t) −e−t^ sin(2t) + 2e−t^ cos(2t) −e−t^ cos(2t) − 2 e−t^ sin(2t)
= −e−^2 t^ sin(2t) cos(2t) − 2 e−^2 t^ sin^2 (2t) + e−^2 t^ sin(2t) cos(2t) − 2 e−^2 t^ cos^2 (2t) = − 2 e−^2 t(sin^2 (2t) + cos^2 (2t)) = − 2 e−^2 t. So, W (y 1 , y 2 )(0) = − 2.
- Find the solution of the following problem:
y′′^ − 6 y′^ + 9y = 0, y(0) = − 1 , y′(0) = 1.
Solution. The characteristic equation is r^2 − 6 r + 9 = (r − 3)^2 = 0, and so there is a repeated root r 1 = 3. Thus, the equation has the general solution y(t) = c 1 ter^1 t^ + c 2 er^1 t^ = c 1 te^3 t^ + c 2 e^3 t.
Applying y(0) = −1 gives c 2 = −1, hence y(t) = c 1 te^3 t^ − e^3 t. As
y′(t) = c 1 e^3 t^ + 3c 1 te^3 t^ − 3 e^3 t,
applying y′(0) = 1 yields c 1 − 3 = 1 hence c 1 = 4. Therefore, the solution of the given initial value problem is y(t) = 4te^3 t^ − e^3 t.
- For the differential equation
(x − 1)y′′^ − xy′^ + y = 0, x > 0 , suppose y 1 (x) = ex^ is a solution. Use the method of reduction of order to find a second solution. Solution. The method of reduction of order begins by guessing that a second solution having the form y(x) = v(x)y 1 (x) = v(x)ex^ also solves the equation. Then we have: y′^ = v′ex^ + vex, y′′^ = v′′ex^ + v′ex^ + v′ex^ + vex = v′′ex^ + 2v′ex^ + vex.
Substituting the above into the original ODE gives:
(x − 1)(v′′ex^ + 2v′ex^ + vex) − x(v′ex^ + vex) + vex^ = 0,
- Determine whether or not the following differential equation has a unique solution. Justify your answer. y′^ =
t^2 + y − 4 , y(2) = 1. Solution. Here we must use the more general Existence and Uniqueness Theorem for 1st Order Equations (Theorem 2. 4 .2), which states that the initial-value problem y′^ = f (t, y), y(t 0 ) = y 0 has a unique solution in a rectangle containing the point (t 0 , y 0 ) only if f and ∂f∂y are continuous on said rectangle. To proceed, we determine the region(s) on which both f and ∂f∂y are continuous, and select the region containing the point (t 0 , y 0 ). So, here f (t, y) =
t^2 + y − 4 , ∂f ∂y (t, y) =
t^2 + y − 4
which are both continuous where y > 4 − t^2 (for f we require t^2 + y − 4 ≥ 0 and for fy we require t^2 + y − 4 > 0). This is the region in the t-y plane above the parabola y = 4 − t^2 , not including the curve itself. Since the point (2, 1) lies in this region, i.e. as 1 > 4 − 22 = 0, by the Existence and Uniqueness Theorem for 1st Order Equations, the given equation has a unique solution y defined on it.
- (a) Find the Wronskian of two solutions y 1 (t) and y 2 (t) for the following equation:
(1 + x^2 )y′′^ + 2xy′^ + y = 0.
(b) Suppose W (y 1 , y 2 )(1) = 3. Find W (y 1 , y 2 )(3). Solution. (a) Since we do not know y 1 (t) and y 2 (t), we compute the Wronskian from the formula in Abel’s Theorem (Theorem 3. 3 .2). Namely, given an equation of the form y′′^ + p(t)y′^ + q(t) = g(t), the Wronskian of two solutions y 1 (t), y 2 (t) is: W (y 1 , y 2 )(t) = ce−^
∫ (^) p(t) dt , where c is a constant to be determined. So, to proceed we first put the given equation in the correct form by dividing through by (1 + x^2 ):
y′′^ + 2 x 1 + x^2 y′^ +
1 + x^2 y = 0.
Then as p(x) = 2 x 1 + x^2
and (^) ∫ p(x) dx =
2 x 1 + x^2 dx = ln(1 + x^2 )
(no absolute value since (1 + x^2 ) is always positive), we obtain:
W (y 1 , y 2 )(x) = ce−^ ln(1+x (^2) ) = c 1 + x^2
(b) If W (y 1 , y 2 )(1) = 3, we have from (a) that c 2 = 3 or c = 6. Thus,
W (y 1 , y 2 )(3) =
1 + (3)^2
- Suppose a student borrows $100, 000 loan to pay tuition. The lender charges interest at an annual rate of 8%. Assume that the interest is compounded continuously and the borrower makes payments continuously at a constant annual rate of $10, 000. How long does it take the student to pay off the debt? Solution. We approach this problem by assuming that the entire balance of the loan is deposited into an account which earns interest at the same rate the lender charges for the loan. Then payments are withdrawals on this account, and the loan is payed off when the account balance reaches $0. The generic formula for this problem is dp dt = rp − k, p(0) = p 0
where p = p(t) is the account balance after t years, r is the lending rate, k is the withdrawal rate, and p 0 is the initial deposit. Here, r = 0.08, k = 10, 000, and p 0 = 100, 000, so we have the initial-value problem: dp dt = 0. 08 p − 10 , 000 , p(0) = 100, 000.
This equation is linear, so writing p′^ − 0. 08 p = − 10 , 000 , p(0) = 100, 000:
- μ(t) = exp
(− 0 .08) dt = e−^0.^08 t.
μ(t)g(t) dt = − 10 , 000
e−^0.^08 t^ dt =
e−^0.^08 t^ = 125, 000 e−^0.^08 t.
- p(t) = e^0.^08 t[125, 000 e−^0.^08 t^ + c] = 125, 000 + ce^0.^08 t.
- p(0) = 100, 000 =⇒ 125 , 000 + c = 100, 000 =⇒ c = − 25 , 000.
- p(t) = 125, 000 − 25 , 000 e^0.^08 t. Now, the student pays off the loan when the account is emptied; i.e. after t years when p(t) = 0. Thus:
0 = 125, 000 − 25 , 000 e^0.^08 t^ =⇒ e^0.^08 t^ = 5 =⇒ t = ln 5
- 08
ln 5 ≈ 20 .12 years.
- The population of a certain mosquito increases at a rate proportional to the current number of mosquitoes. Suppose that the insect population doubles each week. There are 200, 000 mosquitoes in the area initially and predators kill 20, 000 each day. Find the number of mosquitoes at time t. Solution. Consider first the insect population in the absence of predators. Since the
