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I don't have to go to lecture ever again. Thank you so much.
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you rock.
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Test tomorrow, you just saved me
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@zodei 'so ln(u(x))=p(x)', sry that was supposed to be 'ln(u(x))=integral(p(x))'... i was supposed to integrate both sides... the integral of u'(x)/u(x) is ln(u(x)) (due to the chain rule - just derive ln(u(x)) to check this) and i forgot to integrate the right hand side. in the next step i didn't make that mistake because i got u(x)=e^integral(p(x)). i realize these remarks can only be a hint to the actual solution but i still hope they are gonna be helpful.
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@zodei difficult question... divide by tan(x) to obtain the standard form of this 1st order linear ODE: y'+p(x)*y=q(x); p(x)=cot(x), q(x)=csc(x). multiply both sides by an integrating factor u(x) to obtain u(x)*y'+p(x)*u(x)*y=q(x). why? if u(x) is chosen carefully the product rule will simplify this to [1.1] (u(x)*y)'=q(x). for that to work u'(x) has to be p(x)*u(x), so u'(x)/u(x)=p(x), so ln(u(x))=p(x), so u(x)=e^integral(p(x)). use this to calculate u(x). plug it into [1.1]. integrate [1.1].
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hey! How you solve this tan x dy/dx + y = sec x? =D
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@platazar MS Paint?
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PLEASE WHAT'S THE NAME OF THIS SOFTWARE ?
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y go to class for 50 mins when i can learn it in 5 mins, in my pj's
Are professors bothered by you being 10x more concise than they are?
shaunmikex 1 year ago 77
you are def the man. thanks so much.
standaman109 2 years ago 28