# Thread: [SOLVED] Help with understanding differential equations + graphs

1. ## [SOLVED] Help with understanding differential equations + graphs

Hey,
Can someone help me approach this problem correctly please:

Find the sol. of the differential equation
e^(-y)dy/dt+2*cos(t)=0
a)sketch three soltuions corresponding to severeal values of constant of integration, C. Does every value of the consnt of intergration corespond to a solution curve? If not, which Cs do occur?
I graphed 3 solutions, but I don't get the second part.

b)Explain if all solutions have the same domain?
I can answer this if i completely get part a

c) sketch the direction feild assosiated with this equation and superimpose your sketches of the solution curves on the direction feild. (suggestion for sketching direction feild: sketch several points along the line t=0, then at the coresponding points along the lines t=pi/6, pi/3, pi/2, etc.
i dont know how to start part c

Ill appreciate any kind of help; thanks

2. ## ok

THe answers to the diff eq. are called integral curves...and yes each different C represents a different integral curve...

you said you can answer two

so ln(e^(-y)dy/dt))=ln(-2cos(t)) and then -y=-ln(y')+ln(-2cos(t)) ...now call y' c...now pick a value for c..say one when c=1 y=-y=-ln(1)+ln(-2cos(t)) =ln(-2cos(t)) and you graph ln(-2cos(t)) and every point on that line will have a direction of 1...now pick more c's and repeat

3. Originally Posted by coe236
Hey,
Can someone help me approach this problem correctly please:

Find the sol. of the differential equation
e^(-y)dy/dt+2*cos(t)=0
a)sketch three soltuions corresponding to severeal values of constant of integration, C. Does every value of the consnt of intergration corespond to a solution curve? If not, which Cs do occur?
I graphed 3 solutions, but I don't get the second part.

b)Explain if all solutions have the same domain?
I can answer this if i completely get part a

c) sketch the direction feild assosiated with this equation and superimpose your sketches of the solution curves on the direction feild. (suggestion for sketching direction feild: sketch several points along the line t=0, then at the coresponding points along the lines t=pi/6, pi/3, pi/2, etc.
i dont know how to start part c

Ill appreciate any kind of help; thanks
a) $e^{-y} \frac{dy}{dt} + 2 \cos t = 0 \Rightarrow e^{-y} \frac{dy}{dt} = -2 \cos t \Rightarrow \int e^{-y} \, dy = -2 \int \cos t \, dt$.

Therefore $- e^{-y} = -2 \sin (t) + A \Rightarrow y = - \ln (2 \sin (t) - A)$.

So your answer is correct - apart from the fact that you have x instead of t in it - (my A is equal to -C which is OK snce both represent arbitrary constants).

Note that the solution is only defined for $2 \sin (t) + C > 0 \Rightarrow C > - 2 \sin t$. So if you want the domain to be all real numbers, that is, -oo < t < +oo, then you can only have values of C > 2 since $-1 \leq \sin t \leq 1$.

Note however that since $-1 \leq \sin t \leq 1$, there will be no value of t for which $2 \sin (t) + C > 0$ if $C \leq -2$ ......

So the answer is that only values C > -2 are allowed ......

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b) If C > 2 then all solutions will have the same domain. But if $C \leq 2$ then the domain will be restricted so that $2 \sin (t) + C > 0$. What happens in the latter case will depend on the value of C. Eg. if C = 1, then you require $2 \sin (t) + 1 > 0 \Rightarrow \sin t > -\frac{1}{2}$ .......

So all solutions don't have the same domain. The domain depends on the value of C.

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c) You choose a value of t, get the value of y from the solution, and then calculate the value of dy/dt at that point. Draw a little arrow starting at the point that has the same gradient as the value of dy/dt at that point.

Do this for different value of t. Your notes should have examples of how to do this.

4. thankyou mr fantastic, you're the best