Solution to Quiz 8 in Calculus II (MATH 106D) - Winter 2005, Exercises of Calculus

The solution to quiz 8 in the calculus ii (math 106d) course offered in winter 2005. The steps to find the solution of the given initial value problem using the technique of separation of variables and euler's method. The solution to the initial value problem is y = ex3/3 and an estimation of y(0.75) using euler's method with initial point (0, 1) and stepsize ∆x = 0.25 is given.

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MATH 106D - CALCULUS II
WINTER 2005
QUIZ 8
NAME:
Show ALL your work CAREFULLY.
Consider the initial value problem
dy
dx =x2y, y(0) = 1.
(a) Use the technique of separation of variables to find the solution of this
differential equation.
By separating the variables, we have
1
ydy =x2dx.
It follows that
ln |y|=x3
3+C
y=Kex3/3.
Since y(0) = 1,K=1and hence the solution is y=ex3/3.
(b) Use Euler’s method to estimate y(0.75) with initial point (0,1) and
stepsize x=0.25.
The n-thstepintheEulersmethodyieldsthey-coordinate as
yn=f(xn1,y
n1)·x+yn1
where the differential equation is dy
dx =f(x, y).Heref(x, y )=x2y.
y1=f(0,1) ·1
4+1=1;
y2=f1
4,1·1
4+1= 65
64;(now x2is 1
2)
y3=f1
2,65
64·1
4+65
64 =1105
1024 ;
y(0.75) = 1105
1024 .
Date: March 25, 2005.
1

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MATH 106D - CALCULUS II

WINTER 2005

QUIZ 8

NAME:

Show ALL your work CAREFULLY.

Consider the initial value problem

dy

dx

= x

2 y, y(0) = 1.

(a) Use the technique of separation of variables to find the solution of this

differential equation.

By separating the variables, we have ∫ 1

y

dy =

x

2 dx.

It follows that

ln |y| =

x^3

3

+ C

⇒ y = Ke

x^3 / 3 .

Since y(0) = 1, K = 1 and hence the solution is y = e x^3 / 3 .

(b) Use Euler’s method to estimate y(0.75) with initial point (0, 1) and

stepsize ∆x = 0.25.

The n-th step in the Euler’s method yields the y-coordinate as

yn = f (xn− 1 , yn− 1 ) · ∆x + yn− 1

where the differential equation is

dy dx =^ f^ (x, y). Here^ f^ (x, y) =^ x

2 y.

y 1 = f (0, 1) ·

y 2 = f

; (now x 2 is

y 3 = f

⇒ y(0.75) =

Date: March 25, 2005. 1