Austin Cholley Jones MAT 275 ONLINE B Spring 2021
Assignment Section 2.2 Separable Equations due 03/15/2021 at 11:59pm MST
1. (1 point)
Solve the initial value problem:
dy
dx =2y,y(0) = 6
y(x) = .
Solution: Separating variables gives
Zdy
y=Z2dx,
so
ln|y|=2x+C
|y|=eCe2x
y=Ke2x
where K=±eC.
The initial conditon y(0) = 6 gives K=6 and
y=6e2x
Answer(s) submitted:
•6eˆ(2x)
(correct)
Correct Answers:
•6 * 2.71828182845905**(2*x)
2. (1 point)
Solve the initial value problem:
y0(x) = p4y(x) + 32,y(1) = 8
y(x) = .
Solution: Separating variables gives
Zdy
√4y+32 =Zdx,
so
1
2√4y+32 =x+C
√4y+32 =2x+C
The initial conditon y(1) = 8 gives C=6 and √4y+32 =2x+6.
Solving for yyields
y=1
4(2x+6)2−32
Answer(s) submitted:
•xˆ(2)+6x+1
(correct)
Correct Answers:
•((2*x + 6)**2 - 32) / 4
3. (1 point)
Solve the initial value problem:
dy
dx =−0.7
cos(y),y(0) = π
6
y(x) = .
Solution: Separating variables gives
Zcosy dy =Z−0.7dx,
so
siny=−0.7x+C
The initial conditon y(0) = π
6gives C=sin π
6=1
2and
siny=−0.7x+1
2.
Solving for yyields
y=arcsin−0.7x+1
2
Answer(s) submitted:
•arcsin(-.7x+sin((pi /6)))
(correct)
Correct Answers:
•arcsin(-0.7*x + 0.5)
4. (1 point) The differential equation
dy
dx =6x+8
9y2+18y+8
has an implicit general solution of the form F(x,y) = K,where
Kis an arbitary constant.
In fact, because the differential equation is separable, we can
define the solution curve implicitly by a function in the form
F(x,y) = G(x) + H(y) = K.
Find such a solution and then give the related functions re-
quested.
F(x,y) = G(x) + H(y) =
.
Solution: Separating variables gives
Z(9y2+18y+8)dy =Z(6x+8)dx,
so
3y3+9y2+8y=3x2+8x+K
Isolating the arbitrary constant, yields
−(3x2+8x)+(3y3+9y2+8y) = K
1
Answer(s) submitted:
•3yˆ3+9yˆ2+8y-3xˆ2-8x
(correct)
Correct Answers:
•k*((3 * (x**2) + 8 *x ) - (3 * (y**3) + 9 *(y**2) + 8 *y))+m
5. (1 point)
The differential equation
dy
dx =35
y1/5+49x2y1/5
has an implicit general solution of the form F(x,y) = K,where
Kis an arbitrary constant.
In fact, because the differential equation is separable, we can
define the solution curve implicitly by a function in the form
F(x,y) = G(x) + H(y) = K.
Find such a solution and then give the related functions re-
quested.
F(x,y) = G(x) + H(y) =
.
Solution: We factor the denominator to get
dy
dx =35
y1/5(1+49x2),
so
y1/5dy =35
1+49x2dx.
Integrating both sides, yields
5
6y6/5=5arctan(7x) + K.
Isolating the arbitrary constant, yields
−5arctan(7x) + 5
6y6/5=K
Answer(s) submitted:
•5arctan(7x)-(5/6)yˆ(6/5)
(correct)
Correct Answers:
•k( 5*arctan(7 *x) + (-5/6 )*(y**(6 /5 )) ) + m
6. (1 point) Solve the initial value problem:
x2
y2−2
dy
dx =1
2y,y(1) = √3
y(x) = .
Solution: Separating variables gives
Z2y
y2−2dy =Z1
x2dx,
so
ln|y2−2|=−1
x+C
y2−2=K·e−1
x
where K=±eC.
The initial conditon y(1) = √3 gives 1 =Ke−1, so K=eand
y2=e1−1
x+2.
Solving for yyields
y=±qe1−1
x+2
Since y(1) = √3>0, it must be
y=qe1−1
x+2
Answer(s) submitted:
•sqrt(eˆ(-(1/x)+1)+2)
(correct)
Correct Answers:
•sqrt(eˆ(1-1/x)+2)
7. (1 point) Solve the initial value problem:
(ln(y))2dy
dx =x2y,y(1) = e2
y(x) = .
Solution: Separating variables gives
Z(lny)2
ydy =Zx2dx,
so
(lny)3
3=x3
3+C
(lny)3=x3+C
The initial condition y(1) = e2gives 23=1+C, so C=7 and
(lny)3=x3+7.
Solving for yyields
lny=x3+71/3
y=e(x3+7)1/3
Answer(s) submitted:
•eˆ((xˆ(3)+7)ˆ(1/3 ))
(correct)
Correct Answers:
•eˆ((xˆ3+2ˆ3-1)ˆ0.333333333333333)
2
8. (1 point)
Find f(x)if y=f(x)satisfies
dy
dx =65yx12
and the y-intercept of the curve y=f(x)is 5.
f(x) = .
Solution: Separating variables gives
Zdy
ydy =Z65x12dx,
so
ln|y|=5x13 +C
y=Ke5x13
Since the y-intercept of the curve is 5, we have that y(0) = 5.,
so K=5, and
y=f(x) = 5e5x13
Answer(s) submitted:
•5eˆ((5xˆ(13)))
(correct)
Correct Answers:
•5 * exp(5 * (xˆ13 ) )
9. (1 point) Solve the initial value problem:
8x−5ypx2+1dy
dx =0,y(0) = −2
y(x) = .
Solution: Separating variables gives
Zy dy =Z8
5
x
√x2+1dx,
so
y2
2=8
5px2+1+C
y2=16
5px2+1+C
The initial condition y(0) = −2 gives 4 =16
5+C, so C=4
5
and y2=16
5√x2+1+4
5.
Solving for yyields
y=±q16
5px2+1+4
5
Since the initial condition y(0) = −2<0, it must be
y=−q16
5px2+1+4
5
Answer(s) submitted:
•-sqrt((((16sqrt(xˆ(2)+1)+4))/5))
(correct)
Correct Answers:
•- sqrt(3.2*(xˆ2 +1)ˆ.5 -3.2 + -2*-2)
10. (1 point) Find the function y=y(x)(for x>0 ) which
satisfies the separable differential equation
dy
dx =9+18x
xy2;x>0
with the initial condition y(1) = 3.
y=.
Solution: Separating variables gives
Zy2dy =Z9
x+18dx,
so
y3
3=9lnx+18x+C
y3=27lnx+54x+C
The initial condition y(1) = 3 gives 33=54 +C, so C=−27
and y3=27lnx+54x−27.
Solving for yyields
y= (27lnx+54x−27)1/3
Answer(s) submitted:
•3(ln(x)+2x-1)ˆ(1/3 )
(correct)
Correct Answers:
•(3 * 9 * ln(abs(x)) + 3 * 18 * x + -27)**(1/3)
11. (1 point)
Find the solution to the differential equation
dy
dt =0.5(y−150)
if y=45 when t=0.
y=
Solution:
SOLUTION
Separating variables gives
Zdy
y−150 =Z0.5dt,
so
ln|y−150|=0.5t+C.
Solving for y,
y=150 +Ae0.5t,
where A=±eC. The initial condition, y(0) = 45, gives 45 =
150 +A, so A=105, and
y=150 −105e0.5t.
Answer(s) submitted:
•-105eˆ(.5t)+150
(correct)
Correct Answers:
•(45-150)*eˆ(0.5*t) + 150
3
12. (1 point)
Solve the initial value problem
du
dt =e4u+8t,u(0) = 7
u(t) = .
Solution: Separating variables gives
Ze−4udu =Ze8tdt,
so
−1
4e−4u=1
8e8t+C
e−4u=−1
2e8t+C
The initial condition u(0) = 7 gives C=1
2+e−28 and
e−4u=−1
2e8t+1
2+e−28.
Solving for uyields
u=−1
4ln(−1
2e8t+1
2+e−28)
Answer(s) submitted:
•-(((ln(-(1/2)eˆ(8t)+(1/(eˆ(28)))+(1/2))))/4)
(correct)
Correct Answers:
•-ln(exp(-7*4) + (4/8) - (4/8)*exp(8*t))/4
13. (1 point) Find the solution of the initial value problem
dy
dx = (x−6)e−2y,y(6) = ln(6)
y(x) = .
Solution: Separating variables gives
Ze2ydy =Z(x−6)dx,
so
e2y
2=x2
2−6x+C
e2y=x2−12x+C
The initial condition y(6) = ln(6)gives 36 =36 −72 +Cso
C=72 and
e2y=x2−12x+72.
Solving for yyields
y(x) = 1
2ln(x2−12x+72)
Answer(s) submitted:
•(((ln(xˆ(2)-12x+72)))/2)
(correct)
Correct Answers:
•(1/2)*ln((x-6)ˆ2+6ˆ2)
14. (1 point)
The differential equation
dy
dx =cos(x)y2+12y+27
8y+42
has an implicit general solution of the form F(x,y) = K,where
Kis an arbitrary constant.
In fact, because the differential equation is separable, we can
define the solution curve implicitly by a function in the form
F(x,y) = G(x) + H(y) = K.
Find such a solution and then give the related functions re-
quested.
F(x,y) = G(x) + H(y) =
.
Solution: Separating variables gives
Z8y+42
y2+12y+27 dy =Zcosx dx,
Z8y+42
(y+3)(y+9)dy =Zcos x dx,
Performing partial fraction decomposition yields
Z3
y+3+5
y+9dy =Zcosx dx,
so
3ln(|y+3|) + 5ln(|y+9|) = sinx+K
Isolating the constant of integration gives
−sinx+3ln(|y+3|) + 5ln(|y+9|) = K
Answer(s) submitted:
•-sin(x)ln((y+3)ˆ3(y+9)ˆ5)
(incorrect)
Correct Answers:
•k(sin(x) -(3*ln(abs(y+3)) + 5 *ln(abs(y+9)) )) +c
4
15. (1 point)
The differential equation
dy
dx =2+6x+6y+18xy
has an implicit general solution of the form F(x,y) = K,where
Kis an arbitrary constnat.
In fact, because the differential equation is separable, we can
define the solution curve implicitly by a function in the form
F(x,y) = G(x) + H(y) = K.
Find such a solution and then give the related functions re-
quested.
F(x,y) = G(x) + H(y) =
.
Solution: Factoring by grouping yields
dy
dx =2(1+3y) + 6x(1+3y)
dy
dx = (2+6x)(1+3y).
Separating variables gives
Zdy
1+3y=Z(2+6x)dx,
so
1
3ln(1+3y) = 2x+3x2+K
Isolating the constant of integration gives
−(2x+3x2) + 1
3ln(1+3y) = K
Answer(s) submitted:
•2xˆ2+6x-ln(1+3y)
(incorrect)
Correct Answers:
•k(x*(2*2 + 6*x)/(2*2 +6 ) + (- 2 /(3*6 + 2*2*3) )* ln(abs(1 + 3 * y))) +c
16. (1 point)
Find the solution to the differential equation
7du
dt =u2,
subject to the initial conditions u(0) = 5.
u=
Solution:
SOLUTION
Separating variables gives
Z1
u2du =Z1
7dt
or
−1
u=1
7t+C.
The initial condition gives C=−1
5and so, after rearranging
slightly,
u=35
7−5t.
Answer(s) submitted:
•-((35)/(5t-7))
(correct)
Correct Answers:
•7*5/(7 - 5*t)
17. (1 point)
Solve the differential equation
dx
dt =6xlnx
t
Assume x,t>0, and use the initial condition x(1) = 3.
x=
Solution:
SOLUTION
Separating variables gives
dx
xlnx=6
tdt,
so
Zdx
xlnx=Z6
tdt,
and thus
ln|lnx|=6lnt+C.
Exponentiating both sides,
|lnx|=eCe6 lnt=eCt6=At6,
where A=eC. Then, with x(1) = 3, ln 3 =A, so
lnx=ln(3)t6,
or
x=et6ln(3)=eln(3t6)=3t6.
Answer(s) submitted:
•3ˆ(tˆ(6))
(correct)
Correct Answers:
•3ˆ(tˆ(6))
5
18. (1 point) (a) Find the explicit solution of the initial value
problem:
dy
dx =2x
−8y−10 ,y(0) = −8
y(x)=
(b) Determine the interval where the solution is defined.
Interval:
Solution: (a)
We first separate the variables
(−8y−10)dy =2x dx
and then integrate
−4y2−10y=x2+C
Substituting the initial condition yields C=−176 so the implicit
solution is
−4y2−10y=x2−176
or, equivalently,
4y2+10y+x2−176 =0.
Using the quadratic formula yields
y(x) = −10 ±p100 −16(x2−176)
8
y(x) = −5±√729 −4x2
4
Substituting x=0 yields
−5±√729
4=−5±27
4
Since the initial condition is y(0) = −8, we have that the explicit
solution is
y(x) = −5−√729 −4x2
4
(b)
The solution is defined when 729 −4x2>0.
Solving the inequality yields the interval
(−13.5,13.5)
Answer(s) submitted:
•-(((sqrt(-4xˆ(2)+729)+5))/4)
•(0.4)U(4,inf)
(score 0.5)
Correct Answers:
•-(1.25+[sqrt(2916-16*xˆ2)]/8)
•(-13.5,13.5)
19. (1 point) Determine whether each first-order differential
equation is separable, linear, both, or neither.
• choose one
• Separable
• Linear
• Both
• Neither
1. dy
dx +exy=x2y2
• choose one
• Separable
• Linear
• Both
• Neither
2. y+sinx=x3y0
• choose one
• Separable
• Linear
• Both
• Neither
3. lnx−x2y=xy0
• choose one
• Separable
• Linear
• Both
• Neither
4. dy
dx +cosy=tanx
Note: You only have two attempts at this problem.
Answer(s) submitted:
•Neither
•Linear
•Linear
•Neither
(correct)
Correct Answers:
•Neither
•Linear
•Linear
•Neither
20. (1 point) Which of the following are separable differen-
tial equations?
•A. dy
dx =x+y
•B. dy
dx x+x2y=x
•C. y0=2x−3y+1
•D. 2xydx + (x2−1)dy =0
•E. dy
dx =x+y2
6
•F. dy
dx =xy
1+x2
•G. dy
dx =xy
•H. dy
dx +yx2=x2
Answer(s) submitted:
•( D, F, G, H )
(correct)
Correct Answers:
•DFGH
7
