MATH 1005 Study Guide - Midterm Guide: Integrating Factor

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MATH 1005B

Test 1 Solutions

January 31, 2012

[Marks]

1. Solve the initial-value problem 2y=cos(x)

y,y(0) = 2.[6]

Solution:

2yy=cos(x)⇒y2=sin(x)+c⇒y=±sin(x)+c.

y(0) = 2 ⇒2=√c⇒c=4 ⇒y=sin(x)+4.

2. Find the general solution of y=x2+y2

xy .[6]

Solution:

y=x2+y2

xy =x

y+y

x,u=y

x⇒u+xu=1

u+u⇒xu=1

u⇒uu=1

x⇒

2u2=ln|x|+c⇒u=±2ln|x|+k⇒y=±x2ln|x|+k.

3. Find the general solution of xy+3y=x+1.[6]

Solution:

y+3

xy=x+1

x⇒I(x)=eR3

xdx =e3ln|x|=|x|3=±x3,andwemaytakeI(x)=x3.

Then x3y+3x2y=x3+x2⇒(x3y)=x3+x2⇒x3y=x4

4+x3

3+c⇒

y=x

4+1

3+c

x3.

4. Find an integrating factor which makes the equation x+y2+xyy=0exact. Donot[6]

solve the equation.

Solution:

P=x+y2,Q=xy, Py=2y, Qx=y, Py=Qx, so the equation is not exact.

Py−Qx

Q=1

xis a function of xonly, so an integrating factor I(x) exists, determined

by I(x)

I(x)=Py−Qx

Q=1

x,whichgivesI(x)=x.

5. Find the general solution of x2y3+(x3y2+2y)y=0.[6]

Solution:

P=x2y3,Q=x3y2+2y, Py=3x2y2=Qx⇒the equation is exact.

fx=P=x2y3⇒f(x, y)=1

3x3y3+g(y),f

y=Q⇒x3y2+g(y)=x3y2+2y⇒

g(y)=2y⇒g(y)=y2+c⇒f(x, y)=1

3x3y3+y2+c, and the general solution of

the equation is 1

3x3y3+y2=k.

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Document Summary

[6: solve the initial-value problem 2y (cid:1) cos(x) = cos(x) y2 = sin(x) +c y = (cid:1) (cid:1) 2yy y(0) = 2 2 = c c = 4 y = x2 + y2 (cid:1: find the general solution of y sin(x) + 4. sin(x) +c. xy y (cid:1) Solution: (cid:1) u2 = ln|x| + c u = (cid:1) , u = x y y x xy. 2: find the general solution of xy. 2 ln|x| + k y = x (cid:1) = |x|3 = x3, and we may take i(x) = x3. + 3x2y = x3 + x2 (x3y) (cid:1) [6: find an integrating factor which makes the equation x + y2 + xyy (cid:1) P = x + y2, q = xy, py = 2y, qx = y, py (cid:5)= qx, so the equation is not exact. Py qx is a function of x only, so an integrating factor i(x) exists, determined.

Related Questions

Follow the six steps below to solve the following: xy' + 6xy =xe^(5x)

Step 1: REWRITE into standard form (y' + P(x)y + Q(x))

a) How does the given equation differ from standard form?

b) What can be done to put the given equation into standardform?

Step 2: IDENTIFY P(x) and Q(x).

a) Where does the standard form indicate P(x) and Q(x) arelocated?

P(x) =

Q(x) =

Step 3: DETERMINE the integrating factor (e^(?P(x) dx)). PlugP(x) as identified in step 2 into the integrating factor forme^(?P(x) dx) . Hint: Simplify e^(?P(x) dx) if possible

a) What is meant by the integrating factor?

b) If the integrating factor looks like e^(ln (x)) how could itbe simplified?

c) What about e^(1/2 ln (x))?

d) What about e^(-ln (x))?

?P(x) dx =

e^(?P(x) dx) =

Step 4: SUBSTITUTE Q(x) from step 2 and e^(?P(x) dx) from step 3into the solution form ye^(?P(x) dx) = ?Q(x)e^(?P(x) dx) + C.

Step 5: INTEGRATE the right side of the result of step 4. Hint:You may need to simplify and/or rewrite the right side before youattempt to integrate.

Step 6: SOLVE for y. To do this divide both sides by theintegrating factor. Remember that all parts of the right sideincluding the C must be divided by the integrating factor.

Solution is _________________________