Question

What is the difference between Initial and Boundary value problems?
What is the interval of validity for the following Initial Value Problem.
6(dy/dx) -2y=xy^4; y(0)=-2

Answer 1

initial conditions are set along the same point

boundary values describe an interval or range

Answer 2

also, that looks like a burnolli equation with that y^4

Answer 3

it most likely has some constraints for y=0, since you have to divide by y^4 to get ahead in it

Answer 4

\[6y^{-4}y'-2y^{-3}=x\]
z=y^-3 ; z^-1/3 = y

y^-4 = z^-4/3

dz/dy = -3 y^-4
dz = -3 y^-4 dy

\[6z^{-4/3}(\frac{z'}{-3z^{-4/3}})-2z=x\]
\[-2z'-2z=x\]
\[z'+z=-\frac{1}{2}x\]

Answer 5

\[e^xz'+e^xz=\frac{1}{2}xe^x\]
\[e^xz=\int \frac{1}{2}xe^x\ dx\]
\[e^xz=\frac{1}{2}e^x(x-1)\]
\[z=\frac{x-1}{2}=y^{-3}\]

Answer 6

\[y=\sqrt[3]{\frac{2}{x-1}}\]

x cannot be 1

Answer 7

when x=0 we are to the left of 1; so our interval of validity is going to be at most:

(-inf,1)

Answer 8

which im pretty sure i just typed out for the practice since I dont see the asker anywhere lol

Answer 9

since we are not solving for a specific solution other than when x=0; i think i did the math correctly :)

Answer 10

well, i see where i dropped a negative tho; but that doesnt seem to change the outcome too much

Answer 11

@amistre64 can you tell me how did you recognize that it is Bernoulli equation?

Answer 12

the power of y was a give away; normally we want to avoid any y parts and its derivatives raised to a power

Answer 13

OK. So are there any specific steps to remember in order to solve Bernoulli equations?

Answer 14

get the y parts to the same side so its set up like a ordinary linear differential equation; then substitute your leftover ys with an appropriate z

Answer 15

once everything is in terms of z; it is your normal ODE

Answer 16

solve for z thru the regular processes; and then zince z = y^... evaluate the z for the y

Answer 17

to make sure i didnt throw in an error roo bad :)

\[e^xz'+e^xz=-\frac{1}{2}xe^x\]
\[e^xz=-\int \frac{1}{2}xe^xdx\]
\[e^xz=-\frac{1}{2}e^x(x-1)+C\]
\[z=-\frac{1}{2}(x-1)+Ce^{-x}=y^{-3}\]
\[-\frac{(x-1)+Ce^{-x}}{2}=y^{-3}\]
\[-\frac{2}{(x-1)+Ce^{-x}}=y^{3}\]
\[-\sqrt[3]{\frac{2}{(x-1)+Ce^{-x}}}=y\]

when x=0; \[-\sqrt[3]{\frac{2}{(-1)+C}}=-2\]
\[\frac{2}{(-1)+C}=8\]
\[16=-1+C;\ C=17\]

\[-\sqrt[3]{\frac{2}{(x-1)+17e^{-x}}}=y\]

this of course goes bad at"

\[x-1+17e^{-x} = 0\]

which of course can be figured out ...later maybe

Answer 18

i am most likely outa practice since the wolf come up with a slightly altered solution
http://www.wolframalpha.com/input/?i=6y%27-2y%3Dxy%5E4%2C+y%280%29%3D-2