Question

Solve the differential equation dy/dx =xy with the initial condition y(0) = 3. Use the solution to find the actual value of y(2). I found its estimate value to be 9.84. How do I find the actual value?

Answer 1

it is seperable right?

Answer 2

Yes

Answer 3

dy/y = x dx

ln(y) = x^2/2 + C

y = Ce^(x^2/2)

y=3 when x=0

3 = C

y = 3e^(x^2/2)

Answer 4

let x=2 and solve for y

Answer 5

how did you determine the estimate?

Answer 6

initial x = 0
Y(0) = 3
delta y = x y delta x
= (0)(3)(.5) = 0
so y(.5) = 3 + 0 = 3

Y(.5) = 3
delta y = (.5)(3)(.5) = .75
Y(1) = 3 + .75 = 3.75

Y(1) = 3.75
delta y = 1 (3.75)(.5) = 1.875
Y(1.5) = 3.75+1.875 = 5.625

Y(1.5) = 5.625
delta y = 1.5(5.625).5 = 4.22
Y(2) = 5.625 + 4.22 = 9.84

Answer 7

a .5 step ... was curious :)

Answer 8

It was part of a 2 step problem where first you find solve the equation with the step size at 0.5 to estimate y(2) then you have to find the actual value and compare the 2.

Answer 9

But the second part doesn't say anything about the step size.

Answer 10

(0,3): y' = 0: +0/2
(.5,3): y' = 1.5: +1.5/2
(.5,3.75) ... your steps look good

the second part is asking you to solve it outright and not by approximations

Answer 11

a) Use Euler's method with step size of 0.5 and the initial condition of y(0) = 3 to estimate
y(2).
b) Solve the differential equation dy/dx=x y with the initial condition y(0) = 3. Use the
solution to find the actual value of y(2).

Answer 12

right.

Answer 13

by using seperation:\[\frac{dx}{dy}=xy\]
\[\frac1y ~{dy}=x~dx\]
\[\int \frac1y ~{dy}=\int x~dx\]
\[ln(y)=\frac 12 x^2+C\]
\[e^{ln(y)}=e^{\frac 12 x^2+C}\]
\[y=e^{\frac 12 x^2}~e^C\]
\[y=Ce^{\frac 12 x^2}\]

Answer 14

when x=0, y=3, C=3: therefore,
\[\Large y=3e^{\frac{1}{2}x^2}\]

Answer 15

when x=2, y=3e^2

Answer 16

I got 3e^2 or 22.167. Is this right?

Answer 17

And I don't get why the step sizes are different even though the 9.84 is an estimate and the other is an actual.

Answer 18

when we work the steps we introduce an error each time. instead of following the exact curve, it is more like trying to steer a tank.