Question

find the general solution of the
differential equation
dy
dx
= (2 cos x − 3) e(−3x+2 sin x)/6 y5/6

Answer 1

In the implicit form

Answer 2

this is the third time you've posted this lol

Answer 3

i'll do a detailed solution

Answer 4

Yes. I have some problem with it still. but nearly there.

Answer 5

Thanks

Answer 6

so this is the equation right?

\[\frac{dy}{dx} = (2cosx -3)e^{\frac{2sinx - 3x}{6}} y^{\frac{5}{8}}\]

Answer 7

we are going to separate variables

we want anything with a y in it on the LHS

Answer 8

y5/6

Answer 9

so divide by y^(5/6)

\[y^{\frac{-5}{6}}\frac{dy}{dx} = (2cosx -3)e^{\frac{2sinx - 3x}{6}}\]

Answer 10

y is y^5/6

Answer 11

Ok

Answer 12

now we are going to integrate both sides with respect to x
\[\int\limits{y^{\frac{-5}{6}}\frac{dy}{dx}}\text{ }dx = \int\limits{(2cosx -3)e^{\frac{2sinx - 3x}{6}}} \text{ } dx\]

Answer 13

on the left hand side, we have the chain rule going on, because we have dy/dx multiplied by a function of y

in fact this chain rule effect allows us to rewrite that left integral such that we are integrating with respect to y:

\[\int\limits{y^{\frac{-5}{6}}} dy = \int\limits{(2cosx -3)e^{\frac{2sinx - 3x}{6}}} \text{ } dx\]

Answer 14

Ok

Answer 15

lets evaluate the left integral right now:

\[6y^{\frac{1}{6}} = \int\limits{(2cosx -3)e^{\frac{2sinx - 3x}{6}}} \text{ } dx\]

notice i haven't added an arbitrary constant just yet. i could have, but when we evaluate the second integral we will get one, and so its easier just to add it in later

\[ 6y^{\frac{1}{6}} = \int\limits{(2cosx -3)e^{\frac{2sinx - 3x}{6}}} \text{ } dx\]

next we will be evaluating the right integral.

Answer 16

Got it

Answer 17

\[I =\int\limits{(2cosx -3)e^{\frac{2sinx - 3x}{6}}} \text{ } dx\]

we can do this by substitution:
let
\[u = 2sinx - 3x\]
so
\[\frac{du}{dx} = 2cosx - 3\]\[dx = \frac{du}{2cosx - 3}\]

\[I = \int\limits{(2cosx -3)e^{\frac{u}{6}}}dx = \int\limits{(2cosx -3)e^{\frac{u}{6}}}\frac{du}{2cosx - 3}\]

\[I = \int\limits{e^\frac{u}{6}}\text{ }du\]

Answer 18

following?

Answer 19

Ok. So (2cosx -3) gets cancelled out?

Answer 20

What is the method call by which you are doing this? composite rule or substitution method?

Answer 21

it's integration by substitution. if you find it difficult then all it is is practise
so we get:


\[6y^{\frac{1}{6}} = \int\limits{e^\frac{u}{6}}\text{ }du = \frac{1}{6} e^{\frac{u}{6}} + C\]

Answer 22

all we need to do now is solve for y:

tip: multiplying or adding to C by a constant does not really affect it, as we just get another arbitrary constant.
\[6y^{\frac{1}{6}} = \frac{1}{6} e^{\frac{u}{6}} + C\]\[y^{\frac{1}{6}} = \frac{1}{36} e^{\frac{u}{6}} + C\]\[y = (\frac{1}{36} e^{\frac{u}{6}} + C)^6\]

Answer 23

Is this the implicit form or explicit form? It seems explicit.

Answer 24

I have tried to use this to resolve the following initial values but it does not seem right. That is why I posted it again to see if it was correct.

Answer 25

Find the particular solution of the differential equation for which y = 1 when x = 0, and then give this particular solution in explicit form.

Answer 26

hmm it does look a bit explicit. remember to sub back in u:
\[36y = (e^{\frac{-3x + 2sinx}{6}}+C)^6\]

i guess to make it implicit you could square both sides..

if you want another one to practise the techniques here:

\[\frac{dy}{dx} = 2xy^2e^{(x^2)}\]

Answer 27

i just thought, squaring both sides probably not a good idea, it doubles up the curve

Answer 28

With Y = 1 and x= 0 . I am trying to find the value for C. But not convinced by what I get.