Mr Hegarty. Thanks ever so much for this tutorial and find the particular solution of the equation: dy/dx = -3 (Y-2)/ [(2x+1)(x+2)]. You shown me, that when I get to the point: ln Iy-2I = ln I ((x+2)/(2x+1))I + C, you need to sub in the values, x=1, y=4 to find the value of C, before taking the equation any further, as you'll only find the value of A and not C. An Excellent tutorial. Thanks V Much Dominic Gecas
I solved this problem with Trig Substitution: (x=tanθ) & (dx/dθ=sec²θ) & √(x²+1) = secθ ∫(tanθ sec²θ/sec²θ)dθ = ∫tanθ dθ = ln|secθ|+C = ln|√(x²+1)|+C = & ln|siny| = ln|√(x²+1)|+C = siny = e^ln|√(x²+1)|+C = siny = e^ln|√(x²+1)|*e^C = siny = Ce^ln|√(x²+1)| = y =arcsin[C|√(x²+1)|] = & x²+1 > 0, so y =arcsin[ C√(x²+1) ] (I got nervous because our answers went in 2 different directions until the end.) ....but you rock!! Thanks for making this video!!
I remember because this drove me insane one of my calculator apps said -ln|cosx| while the other said ln|secx| (not that either was wrong, but I always trusted my brain after that.)
Why is it when you take the exponential of ln sin y it cancels the ln leaving sin y, however, when taking it for ln 1+x^2 it doesnt. Why are you left with a e term for it?
Thank you for these grade-saving videos!
THANKYOU VERY MUCH YOU SAVED MY LYF
Mr Hegarty. Thanks ever so much for this tutorial and find the particular solution of the equation: dy/dx = -3 (Y-2)/ [(2x+1)(x+2)]. You shown me, that when I get to the point:
ln Iy-2I = ln I ((x+2)/(2x+1))I + C, you need to sub in the values, x=1, y=4 to find the value of C, before taking the equation any further, as you'll only find the value of A and not C.
An Excellent tutorial.
Thanks V Much
Dominic Gecas
Thank you so much. It helped me a lot.
I solved this problem with Trig Substitution:
(x=tanθ) & (dx/dθ=sec²θ)
& √(x²+1) = secθ
∫(tanθ sec²θ/sec²θ)dθ =
∫tanθ dθ =
ln|secθ|+C =
ln|√(x²+1)|+C =
& ln|siny| = ln|√(x²+1)|+C =
siny = e^ln|√(x²+1)|+C =
siny = e^ln|√(x²+1)|*e^C =
siny = Ce^ln|√(x²+1)| =
y =arcsin[C|√(x²+1)|] =
& x²+1 > 0, so
y =arcsin[ C√(x²+1) ]
(I got nervous because our answers went in 2 different directions until the end.)
....but you rock!! Thanks for making this video!!
Thankyou! This has helped me a lot :)
Much appreciated mate
thanku🥰
thanks alot!
Very good video, although I hate it when you say ln
Great video, really helpful (apart from the fact you say "lunn" not natural log or l n hahah)
Why don't we add a constant with y after integrating? (In example 3)
isnt the integral of tanx equall to -ln(cosx)
Yes.
same thing: ln|(secx)| = -ln|(cosx)|
= ln|(cosx)^(-1)|
= ln|(1/cosx)|
= ln|(secx)|
I remember because this drove me insane
one of my calculator apps said -ln|cosx| while the other said ln|secx|
(not that either was wrong, but I always trusted my brain after that.)
Thanks! do you do private tutorials?
Why is it when you take the exponential of ln sin y it cancels the ln leaving sin y, however, when taking it for ln 1+x^2 it doesnt. Why are you left with a e term for it?
M LV that constant is irritating isn't it? You have to separate the constant before simplifying
integration of 2/2x+1 should be 2ln2x+1 right?
Monica Chandwani no just ln2x+1
Take 2 out of the bracket, so you are integrating 1/(2x+1). This integrates to 1/2 ln2x+1, hence times the 2 out of the diffrential to leave ln2x+1
is arcsin the same as pressing shift + sin on your calculator.
yeah, so it is
Lol Love how the original comment got no likes, but the reply did