Fun Fun Fun. Find the general solution of the given differential equation.Give the largest interval I over which the general solution is defined. Determine whether there are any transient terms in the general solution. cos(x)y'+(sin(x))y=1
\[e^{\int\limits{}_{}p(x)dx}=e^{\int\limits{}_{}\tan(x)dx}\]
=\[e^{-\ln(\cos(x)+c}\]
=\[=e^{\ln(\cos(x^{-1})+c}\]
\[=\cos(x^{-1})+c\]
\[\int\limits_{}^{}\frac{d}{dx}[\cos(x^{-1})y]=\int\limits_{}^{}\frac{1}{\cos(x)}*\cos(x^{-1})dx\]
\[\cos(x^{-1})y=\int\limits{}_{}\frac{\cos(x^{-1}}{\cos(x)}dx\]
\[e ^{\int\limits \tan(x)} = e ^{-\ln \cos(x)} = e ^{\ln {1 \over \cos(x)}} = \sec(x)\]
aha not simplified enough alright
\[\sec(x)y=\int\limits_{}^{}\sec^2(x)dx\]
\[\sec(x)y=\tan(x)+c\]
so far soo good?
yeah the book just multiplies by cos(x) to get y=sin(x)+c(cos(x) book makes no sense at all -.- leaves it in c= and then sometimes y=
because sec(x) = 1/cos(x), they multiply both sides by cos(x) to isolate y
but yes, looks good Outkast3r09
Looks good to me. This one is easy to solve for y and its usually best to have an explicit solution
yeah i undrstand that but sometimes it will say differential general and still leave it as c= anyways i was away for cases could you expain how this works jim"?
not sure what you mean
well in this equation ther eis only one case but sometimes you can get two cases such as case 1: x cannot equal 0 case 2: x = 0
it has to do when you divide by something just didn't understand it the way he explained i'm sure i'd understand if he had explained it
hmm well I see no domain issues with y=sin(x)+c*cos(x), so you don't have to worry about cases there...but...you may have to worry about cases in intermediate steps. I'm not too sure.
trying to find a case problem hold on
alright, I'm looking up dealing with cases, but not finding much
yeah the book doesn't explain it as much but my teacher is a Dr in Differential Equations and always checks every case.. most of the time it's just the general but i've seen problems where there was 2 answers
hmm alright, let me look some more
You might be on to something when you say y(0) So y(0)=sin(0)+c*cos(0) ---> y(0) = c And for nonzero x, y(x)=sin(x)+c*cos(x) Another case is x = kpi/2 where k is some integer. If this is the case, then y(x) = (-1)^(k+1) Not sure if I thought of every possible case though.
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