Question

Find the value of A so that the equation y'-x^6y-x^6=0 has a solution of the form y(x)=A=Be^(x7/7) for any constant B.

wat.

Answer 1

*A+Be^(x^7/7)

Answer 2

multiply the equation thru by IF :

\(\large e^{-\frac{x^7}{7}}\)

Answer 3

first two terms reduce to form (fg)'
which is easily integrable to fg

Answer 4

thanks. :>

Answer 5

wat did u get for final solution ha ?

Answer 6

-1!

Answer 7

i knw u gave up lol

Answer 8

uhhh i didnt actually. I didnt get the way you were explaining though so i just did it the original way i was doing and found out i had just made a calculation error the first time around.

Answer 9

thats fine :)
may i knw the final solution u got ?

Answer 10

I got A = -1.

Answer 11

Okay, could you shine some light on the "original way" you talking about ?

The only way I ever learned about solving these was by multiplying IF (integrating factor) :
\( e^{\int p dx} \)

It converts y' and y terms to \((fg)'\) form, so that we can use the fundamental theorem calculus :
\(\int (fg)' = fg\)

Answer 12

Here is the solution to present problem using Integrating Factor method :

\(\large y'-x^6y-x^6=0 \)

multiply the IF \(\large e^{\frac{-x^7}{7}}\) thru out :

\(\large e^{\frac{-x^7}{7}}y' -e^{\frac{-x^7}{7}} x^6y - e^{\frac{-x^7}{7}} x^6 = 0\)

\(\large e^{\frac{-x^7}{7}}y' -e^{\frac{-x^7}{7}} x^6y = e^{\frac{-x^7}{7}} x^6 = 0\)

\(\large (e^{\frac{-x^7}{7}}y)' = e^{\frac{-x^7}{7}} x^6 \)

Integrating both sides :

\(\large \int (e^{\frac{-x^7}{7}}y)' dx = \int e^{\frac{-x^7}{7}} x^6 dx\)

\(\large e^{\frac{-x^7}{7}}y = \int e^{\frac{-x^7}{7}} x^6 dx\)

right side is trivial to integrate

Answer 13

\(\large e^{\frac{-x^7}{7}}y = \int e^{\frac{-x^7}{7}} x^6 dx \)

\(\large e^{\frac{-x^7}{7}}y = -e^{\frac{-x^7}{7}} + c \)

\(\large y = -1 + ce^{\frac{x^7}{7}} \)

Answer 14

^^ gives -1 as well, but for some reason i dont see how we can do this in any other way :o

Answer 15

I just took the derivative of y, subtracted x^6 time y minus x^6 and ended up with this \[Bx^6e^\frac{ x^7 }{ ^7 }-x^6(A+Be ^\frac{ x ^7}{ ^7 })-x^6\] Then i factored out x^6 and was able to cancel Be^x^7/7 with -Be^x^7/7
So i was just left with -A-1=0
-A=1
A=-1

Answer 16

Awesome ! I see u worked it backwards ! thats a cool trick :)

Answer 17

woooo! :) thanks for explaining it an additional way.

Answer 18

:) i see a third simpler way lol

Answer 19

but i guess wat you did is the simplest way possible !

Answer 20

I'll put the third method anyways :-

Answer 21

Method 3 :-

\(y'-x^6y-x^6=0 \)

\(y' -x^6(y-1) = 0\)

\(y' = x^6(y-1)\)

\(\large \frac{dy}{dx} = x^6(y-1)\)

\(\large \frac{dy}{y-1} = x^6dx\)

Integrating both sides :

\(\large \int \frac{dy}{y-1} = \int x^6dx\)

\(\large \ln (y-1) = \frac{x^7}{7} + c\)

\(\large y-1 = Be^{\frac{x^7}{7}}\)

Answer 22

this method may feel easy to work out..

Answer 23

*typo
it should be \((y+1)\) all the way