Question

i need help with this problem Find the particular solution of the differential equation (dy/dx) +4y=8 satisfying the initial condition y(0)= 0

Answer 1

hi can u help plz

Answer 2

i got y=-2e^(4x)+2 but its wrong can some one show me where the wrong

Answer 3

how u get 4xy

Answer 4

integral

Answer 5

integral of what

Answer 6

im a bit rusty, but that should b it.

Answer 7

its wrong
coz i tried it on the web work it say is not correct

Answer 8

coz its online homework

Answer 9

dy/dx = 8 - 4y
dy / (8-4y) = dx

u = 8 - 4y
du/dy = -4
du/-4 = dy

du / -4u = dx

(lnu)/-4 = x + C

ln(8-4y) = -4x + C

Ce^-4x = 8 - 4y
y = 2 - Ce^-4x
y(0) = 2 - C = 0
C = 2

y = 2- 2e^-4x

verification :

dy/dx = 8e^-4x
8e^-4x = 8 - 4y ?
8e^-4x = 8 -4(2-2e^-4x)
8e^-4x = 8e^-4x

Answer 10

thanks its correct now

Answer 11

We have that\[\frac{dy}{dx}+4y=8.\]This is a first-order, linear ODE. As such, it can be solved using an integrating factor.\[\mu(x)=e^{\int4dx}=e^{4x}.\]Now you multiply the equation by this integrating factor to obtain\[e^{4x}\frac{dy}{dx}+4e^{4x}y=8e^{4x}.\]The LHS of this equation is the expansion of the product rule\[\frac{d}{dx}[e^{4x}y]=8e^{4x}.\]Integrating, we see\[e^{4x}y=8\cdot\frac{1}{4}e^{4x}+c=2e^{4x}+c.\]Solving for y\[y=2+\frac{c}{e^{4x}}.\]Using the initial value\[0=2+\frac{c}{e^{4\cdot0}},\]\[0=2+c,\]\[c=-2.\]Therefore,\[y=2-\frac{2}{e^{4x}}.\]

Answer 12

Nevermind; too late. ;P

Answer 13

ok thats right but can u help me with this one too

Answer 14

yet much better lol

Answer 15

this is a good page for these: http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx

Answer 16

Solve the separable differential equation (dy/dx)=(-0.7/cosy) and find the particular solution satisfying the initial condition
y(0)=pi/3

Answer 17

thanks that good page i ll red it

Answer 18

i got sin y=-0.7x +0.866 but i need to solve for y not sin y

Answer 19

just do inverse sin both sides

Answer 20

cosydy = -0.7dx
siny = -0.7x + C
y = arcsin(-0.7x + C)
arcsin(C) = pi/3
C = sin(pi/3)
C = (3^0.5)/2

y = arcsin(-0.7x + (3^0.5)/2)

Answer 21

ok

Answer 22

so jus the arc of it will be fine

Answer 23

i didnt know that thanks :)

Answer 24

what about this q: Find the particular solution of the differential equation )x^2dy)/(y^2-3dx)=(1/2y) satisfying the initial condition y(1)=(4)^(1/2)

Answer 25

are those dy's and dx's in the right place?

Answer 26

ya

Answer 27

if it's not supposed to be x^2/(y^2-3)(dy/dx)=1/2y I didn't even know you could write something like that. And why are the conditions written as y(1)=4^(1/2) and not just y(1)=2 ?
I think I need to appeal to someone greater than I.

Answer 28

yes that right

Answer 29

You crack me up TuringTest, in a good way. :P <3

Answer 30

tell me across, does the expression we were given look doable?

Answer 31

I gave it a second look and was perplexed to see a dx thrown in the denominator with no proportional connection to dy whatsoever. He must have typoed it.

Answer 32

talarowen, make sure you write the problems exactly as they're given if you ever want us to help you solve them. :p