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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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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MATH 2930 - Worksheet 2 Solutions