Question

HELP!!! VERY URGENT!!! Consider the differential equation given by dy/dx=xy/3. Find the particular solution y=f(x) to the given differential equation with the initial condition f(0) = 4.

Answer 1

separate the variables

Answer 2

\[\frac{dy}{dx} = \frac{1}{3}x y\]Separate the variables:
\[y^{-1} dy = \frac{1}{3} x dx\]Integrate both sides\[\ln y = \frac{1}{6}x^2 + C\]Take the anti-log of both sides
\[y = e^{\frac{1}{6}x^2 + C}\]Now set \(y = 4, x = 0\) and find the value of \(C\).

Answer 3

before finding C, rewrite it as:

\[y = Ce^{\frac{ x^2 }{ 6 }}\]

Answer 4

need tinier font ... i can almost make it out

Answer 5

lol

Answer 6

\[\tiny{y=e^{\frac{1}{6}x^2+C}}\]
Better? :-)

Answer 7

lol

Answer 8

much more unreadable, thnx ;)

Answer 9

you're welcome! :-)

Answer 10

Thank you so much everyone! i really appreciate the help! However, should my final answer be \[y=e^{\frac{ 1 }{ 6 }x^{2}+C}\]

Answer 11

I really appreciate the help!*

Answer 12

close: you need to find the value of C that makes y = 4 when x = 0

Answer 13

as suggested by @euler271, it may be easier if you rewrite as \[\large{ y = C e^{\frac{1}{6}x^2}}\] before plugging in the initial conditions.

Answer 14

you can do it my way, too, it just takes an extra step to combine the resulting exponent into a nicer form.

Answer 15

I made the font larger this time for @amistre64 but he/she's already left :-)

Answer 16

I'm kind of stuck trying to solver for \[C\]

Answer 17

\[\Huge \color{#05ad00}{ y = C e^{\frac{1}{6}x^2}}\]

Answer 18

\[\huge{4 = C e^{\frac{1}{6}(0)^2}}\]\[\huge{4 = Ce^0}\]

Answer 19

\[\Huge \color{#05ad00}{ y = C e^{\frac{1}{6}x^2}}~:~x=0, y=4\]
\[\Huge { 4 = C e^{\frac{1}{6}0^2}}\]
\[\Huge { 4 = C e^{0}}\]
\[\Huge { 4 = C}\]

Answer 20

THANK YOU SO MUCH EVERYONE! I REALLY APPRECIATE IT!

Answer 21

Had you not rewritten before finding C, you would get a different value for C:
\[\large{4 = e^{\frac{1}{6}(0)^2+C}}\]\[4 = e^C\]\[C = \ln 4\]Then your solution would be \[\large{y = e^{\frac{1}{6}x^2+\ln 4} = e^{\frac{1}{6}x^2}*e^{\ln 4} }\]but of course \[e^{\ln 4} = 4\]so that becomes \[\large{y = 4e^{\frac{1}{6}x^2}}\]just like the other way.