Solving Lcos(y)y' + sin(y) = 0

[projection]

Homework Statement

Lcos(y)y' + sin(y) = 0

The Attempt at a Solution

I separated them into their respective M and N functions. took their integrals:

\int Lcos(y)dy = Lsin(y) + g(y) + C

\int sin(y)dx = -cos(y) + f(y) + C

there the solution would be Lsine(y) -cos(y) + C.

this is confusing me a bit as I am used to seeing commonality between the two equations and the added f(y) and g(x) take form of the differences to create 2 similar equations... am i doing this right or am i missing something?

 

[projection]

opps, i think the L is actually a t... so it is not a constant i think...

so would this be the solution now?
\int tcos(y)dy = tsin(y) + g(y) + C

\int sin(y)dt = \int dt= tsin(y) + f(y) + C

and the solution is tsin(y) ?

 

[rock.freak667]

If y' means dy/dt then you have

N=tcos(y) and M=sin(y). If we take your first integral we have

∂f/∂y= N = tcos(y) or f(x,y) = tsin(y) + g(t)

Now you also know that ∂f/∂t = M. So you need to find ∂f/∂t and equate it to M to get g(t).

 

[LeonhardEuler]

Yes, the equation was not even an exact equation the way you had it before. I'm assuming y' means dy/dt.

Of course, tsin(y) is not the solution to the equation, but the function whose total derivative gives the differential equation.

 

[projection]

Yes, the equation was not even an exact equation the way you had it before. I'm assuming y' means dy/dt.

it doesn't say, but first i thought it was a constant, but remembered than most constants in DEs come as lamba or phi etc... so looking closer at the text and removing smudge made me see it was a t...

so have i done it right? the solution is tsin(y)... hate when text don't have answers for even numbered questions.

 

[LeonhardEuler]

it doesn't say, but first i thought it was a constant, but remembered than most constants in DEs come as lamba or phi etc... so looking closer at the text and removing smudge made me see it was a t...

so have i done it right? the solution is tsin(y)... hate when text don't have answers for even numbered questions.

It depends what you were looking for. If you were looking for F(y,t) such that dF(y,t) = 0 gives the equation, then yes. If you were looking for y(t) that solves the DE, then there is one more step to go.

 

[projection]

It depends what you were looking for. If you were looking for F(y,t) such that dF(y,t) = 0 gives the equation, then yes. If you were looking for y(t) that solves the DE, then there is one more step to go.

the exact problems i have been doing usually stop at this step... and only continue if intial conditions are provided... since no intial value are given (just says test exactness and solve), i am unsure about what more i must do.

 

[LeonhardEuler]

When you solve an exact DE, you take an equation of the form I(y,t)dy + J(y,t)dt = 0 and show that it is equivalent to an equation dF(y,t) = 0, so that
F(y,t) = C.
The solution to the original equation can be written in implicit form as above, or explicitly by solving for y as a function of t. As it stands t sin(y) is not a solution,
t sin(y) = C
is, and to get an explicit form you need to solve for y as a function of t. It depends on what the exact directions are.

 

[SammyS]

Homework Statement

t·cos(y)·y' + sin(y) = 0 (corrected per post #2)

The Attempt at a Solution

...

It is separable:

t\,{{dy}\over{dt}}=\,-\,{{\sin(y)}\over{\cos(y)}}

-\,{{\cos(y)}\over{\sin(y)}}\,dy={{dt}\over{t}}

Integrate both sides. (I prefer to say, "Take the anti-differential.")

 

[projection]

checking my notes, the example solution stops at the statement that "the equation (exe: tsin(y)) determines y(t) implicity"... so in this case, the solution would be tsin(y) = c (implicit form)

so, unless i solve for y as a function of the other variable, i should state the solution is an implicit definition?

 

[LeonhardEuler]

checking my notes, the example solution stops at the statement that "the equation (exe: tsin(y)) determines y(t) implicity"... so in this case, the solution would be tsin(y) = c (implicit form)

so, unless i solve for y as a function of the other variable, i should state the solution is an implicit definition?

Yes.

 

[projection]

Yes.

thank you.