ENGR 213 Study Guide - Final Guide: Wronskian, University Of Manchester, Damping Ratio
31 May 2018
School
Department
Course
Professor
Solutions to the Final exam of Winter 2009
Applied Ordinary Differential Equations
ENGR 213
(1) (a) This is a separable equation which can be re-written in the form
dy
(2y+ 3)2=dx
(4x+ 5)2.
To integrate the left hand side, use the substitution u= 2y+ 3, thus du = 2dy,
while, for the right hand side, use w= 4x+ 5, dw = 4dx. Therefore,
Zdy
(2y+ 3)2=1
2Z1
u2du =−1
2u+c=−1
2(2y+ 3) +c,
Zdy
(4x+ 5)2=1
4Z1
w2dw =−1
4w+c0=−1
4(4x+ 5) +c0.
Using a single constant Cto replace the difference c0−c, we have the general
solution of the equation in implicit form:
1
2(2y+ 3) =1
4(4x+ 5) +C.
(b) This ODE is linear, hence we shall first put it in its standard form (we’ll solve it
for x > 0):
dy
dx +1
xy=1
xex.
Thus P(x) = 1/x and Q(x) = ex/x and we look for an integrating factor (hence in
the following we choose the constant of integration zero)
µ(x) = eR1
xdx =eln x=x.
We multiply the equation with µto obtain xy0+y=exor (xy)0=ex. Integrating
both sides, we obtain
xy =ex+C, C = constant,⇒y(x) = ex
x+C
x, C = constant,
the general solution in explicit form.
We now make use of the initial condition y(1) = 2 to determine the constant:
y(1) = e+C= 2 ⇒C= 2 −e⇒y(x) = ex
x+2−e
x.
(2) By denoting M(x, y) = y2cos x−3x2y−2x, N(x, y)=2ysin x−x3+ ln y, we check
that ∂M
∂y (x, y) = ∂N
∂x (x, y) = 2ycos x−3x2
hence the equation is exact.
1
2
We therefore start to look for a function f(x, y) whose partial derivatives with respect
to x, respectively y, are Mand respectively N. We do so by integrating M(x, y) with
respect to x:
f(x, y) = y2sin x−x3y−x2+C(y).
We then differentiate fwith respect to yand, setting the result equal to N, obtain
2ysin x−x3+C0(y) = ln y.
Thus C(y) is an antiderivative of ln y. We can find it by integration by parts (u=
ln y, dv =dy):
C(y) = Zln y dy =yln y−Z1dy =yln y−y+c, c = constant.
Hence f(x, y) = y2sin x−x3y−x2+yln y−y+cand the solution to the ODE (in
implicit form) is
y2sin x−x3y−x2+yln y−y+c= 0, c = constant.
(3) We’ll use the substitution u=x+y⇒y=u−x, then dy
dx =du
dx −1. Therefore, the
substitution leads to the related ODE
du
dx −1 = tan2u⇔du
dx = 1 + tan2u.
As tan u=sin u
cos u, we note that 1 + tan2u=sin2u+ cos2u
cos2u=1
cos2u.Then
du
dx =1
cos2u⇒cos2u du =dx
whose solution is Z1 + cos(2u)
2du =x+c⇒u
2+sin(2u)
4=x+c, thus the
general solution of the original ODE (in implicit form) is
x+y
2+sin(2(x+y))
4=x+c, c = constant.
(4) For the first 4 seconds, the velocity is constant (equal to 100 m/sec), then the velocity
is a solution of the ODE dV
dt =−0.002V2.
This ODE has the solution
−1
V=−0.002t+c, V (0) = 100
or
V(t) = 1
0.002t+ 0.01 ⇒1
0.002t+ 0.01 = 20 ⇒t= 20.
Document Summary
Solutions to the final exam of winter 2009. Engr 213 (1) (a) this is a separable equation which can be re-written in the form dy (2y + 3)2 = dx (4x + 5)2 . To integrate the left hand side, use the substitution u = 2y + 3, thus du = 2dy, while, for the right hand side, use w = 4x + 5, dw = 4dx. Using a single constant c to replace the di erence c c, we have the general solution of the equation in implicit form: + c. (b) this ode is linear, hence we shall rst put it in its standard form (we"ll solve it for x > 0): dy dx. Thus p (x) = 1/x and q(x) = ex/x and we look for an integrating factor (hence in the following we choose the constant of integration zero) (x) = er 1 x dx = eln x = x.
