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8/4/2019 MATH 2930 - Worksheet 2 Solutions
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MA 2930, Feb 2, 2011
Worksheet 2 Solutions
1.
For each of the following differential equations identify (a) its order, (b)whether its linear or non-linear, and (c) among the methods youve learnedso far which you can use to solve it if any:
(1) y = y2
(2) y = y + t2
(3) y = y + t(4) y2 = y cos t
(5) sin t y = y2 cos t
Recall that the orderof a differential equation is the order of the highest-order derivative of the dependent variable present in it.
Also recall that the equation is linear if it is linear in the dependentvariable and all its derivatives; otherwise it is called non-linear.
So far we have seen two methods of solution:(a) Use of an integration factor if the equation is first-order linear(b) Separation of variables if the equation is first order and can be put in
the form M(x)dx = N(y)dyNow it is clear that the eq.(1) is first-order, but non-linear and in fact separable (y2dy = dt)(2) is first-order linear and can be solved using the integrating factor et;(3) is second-order linear; we dont yet have a method for such an equa-
tion;(4) is first-order non-linear, but can not be put in the form of a separable
equation, though an ad-hoc method may apply;(5) is first-order non-linear and separable (y2dy = cot tdt)
2.
Solve the following differential equations, describe how the long-term (i.e.,as t ) behavior of the solutions depends on the initial conditions, andwhere the solutions achieve their maximum and minimum values if any:
(1) y = 2y2 + xy2
(2) ty + 2y = sin t
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(3) (sin t)y + (cos t)y = et
(4) y2
(1 x2
)1/2
y
= arcsin x(5) x3y + 4x2y = ex
To solve these equations well first identify what kind of an equation it isand then choose an appropriate method.
(1) The equation is first-order, but non-linear, so we ask if it is separable.It turns out that it is:
y = (2 + x)y2
y2y = (2 + x)y2dy =
2 + x dx
y1 = 2x + x2/2 + cy = 24x+x2+c
Since initial condition determines the integration constant, we need to showhow limx y depends upon c. But as x , y 0 irrespective of c,and therefore, of any initial condition. To find its extrema we use the usualmethod which I leave to you.
(2) The equation is first-order, linear, so we need to find an integrationfactor. First we put the equation in the standard form, y + p(t)y = g(t):
y +2
ty =
sin t
t
Then, an integration factor is exp(
p(t)dt) = exp(
2t dt) = exp(2ln |t|) =
exp(ln t2) = t2. Multiplying the equation by it we get
d
dt(t2y) = t sin t
t2y =
t sin t dt
t2y = t cos t + sin t + cy = t1 cos t + t2 sin t + ct2
As t , y 0 regardless ofc, and thus, regardless of any initial condition.
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(3) The equation is first-order, linear, so we need to find an integration
factor. In order to do so we put the equation in the standard form:
y + cot ty = et csc t
Thus, an integration factor is exp(
cot tdt) = exp( ln csc t) = exp(ln sin t) =sin t. Multiplying the equation by it we get
d
dt(sin ty) = et
sin ty = et + c
y = (et + c)csc t
As t , y takes larger and larger swings over any interval of 2, the periodof csc t. Therefore, y does not have a limit regardless of c.
(4) The equation is first-order non-linear, but it is separable:
y2(1 x2)1/2y = arcsin xy2dy =
arcsin x(1x2)
dx
The x-integral can be integrated after the substitution u = arcsin x (forexample) to yield
y3
3= (arcsinx)
2
2+ c
y = (32
(arcsin x)2 + c)1/3
Since arcsin x, and therefore y, is defined only on the interval [-1,1], lettingx doesnt make sense.
(5) The equation is first-order, linear, so we need to find an integrationfactor. In order to do so we put the equation in the standard form:
y + 4x1y = x3ex
Thus, an integration factor is exp(
4x
dx) = exp(4 ln |x|) = exp(ln x4) = x4.
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8/4/2019 MATH 2930 - Worksheet 2 Solutions
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Multiplying the equation by it we get
d
dx(x4y) = xex
x4y =
xex dx
x4y = xex ex + cy = x3ex x4ex + cx4
As x , y 0 regardless ofc, and thus, regardless of any initial condition.
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