I'm thinking about the "+ C" and how it feels almost cosmetic. Here's how I make it feel important. In differential equations class, we eventually get around to learning about general and particular solutions. Let's introduce the idea much earlier, like in early days of antiderivatives. So consider: dy/dx = x The complete solution to "y" will be the general solution ("y_g") plus the particular solution ("y_p"). To find the general solution, first solve dy/dx = 0. The solution to that is "y_g = C". You'll always get "y_g = C" if what's on the left is "dy/dx". Now we find the particular solution to "dy/dx = x"; we get "y _p= x^2 / 2". Add the particular and general solutions, and you get "y = x^2 / 2 + C". We got the complete answer, not by slapping an arbitrary "+ C" on there, but because it actually belongs. Same approach would apply if the initial problem were "dy/dx + y = x". First we'd need to solve "dy/dx + y = 0", and it WOULDN'T be "y_g = C", but rather "y_g = C*e^(-x)". The particular solution would be "y_p = x - 1", leading to a total solution of "y = x - 1 + C*e^(-x)". Same methodology, very few arbitrary steps.
Why not using a hyperbolic substitution? In this case, x = sinh(t) and dx = cosh(t) dt 🙂 Then use cosh²(t) - sinh²(t) = 1, so the integrand simplifies to cosh²(t) (or -cosh²(t)) which is easily integrated to t/2 + 1/2 sinh(t) cosh(t) + c or 1/2 arsinh(x) + 1/2 x sqrt(x² + 1) + c after back-substitution (and -t/2 - 1/2 sinh(t) cosh(t) + c or -1/2 arsinh(x) - 1/2 x sqrt(x² + 1) + c for the other case). Learned to appreciate hyperbolic substitutions after reading this great book: Khristo Boyadzhiev, Special Techniques For Solving Integrals: Examples And Problems 🙂
I play with ODEs like (1-x^2)y'' - xy' + n^2y=0 (1-x^2)y'' - 2xy' + n(n+1)y=0 and problem for me is not solving it with power series but with condition y(1)=1
problem Your two methods have different answers. How come there's a negative in front of the ln in the answer for method 1, but not in the answer for method 2? (dy/dx)² - 1 = x² (dy/dx)² = x² + 1 dy/dx = ±√(x² + 1) ∫ dy = ±∫ √(x² + 1) dx y = ±∫ √(x² + 1) dx Solution is for the positive branch and the negative will also be a solution. The integral will be completed and the constant of integration added to the right side at the end. y = ∫ √(x² + 1) dx Let x = tan t t = tan⁻¹ x dx = sec² t dt y = ∫ sec³ t dt = ∫ (1 + tan²t) sec t dt = ∫ sec t dt + ∫ tan²t sec t dt = ∫ sec t dt + ∫ sin²t /(cos³ t ) dt Let r = cos t √(1-r²) = sin t -ain t dt = dr -dr = sin t dt Note that dr has one of the sin t terms of sin²t and the other sin t is in the √(1-r²). y = ∫ sec t dt - ∫ √(1-r²)/(r³) dr Integrate ∫ √(1-r²)/(r³) dr by parts. dv = dr /(r³) u = √(1-r²) du = -r dr /√(1-r²) v = -(½) 1/(r²) v du = dr /[2 r√(1-r²) ] uv. = -(½) √(1-r²)/(r²) y = ∫ sec t dt - [ uv - ∫ v du ] = ∫ sec t dt - [ -(½) √(1-r²)/(r²) - ∫ dr /[2 r√(1-r²) ]] = ∫ sec t dt + √(1-r²)/(2r²) + ∫ dr /[2 r√(1-r²) ]] Replace r = cos t y = ∫ sec t dt +√(1-r²)/(2r²) - (½) ∫ sec t dt y = (½) ∫ sec t dt + (1/2) tan t sec t We know ∫ sec t dt = ln | sec t + tan t |. y = (1/2) ln | sec t + tan t |+ (1/2) tan t sec t t = tan⁻¹ x x = tan t √(x²+1) = sec t 2y = ln (√(1+x²) + x)+ x√(1+x²) y = (½) ln (√(1+x²) + x )+ (½) x√(1+x²) + c , where c is a constant of integration. answer y = ± ½ [ln (√(1+x²) + x ) + x√(1+x²) ] + c, c ∈ ℝ
I don't understand differential equations. No one has ever explained the concept clearly. And this video didn't help at all. I completed a bachelor's degree in mathematics but lost grades due to these. The lecturers were also useless at explaining these.
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I'm thinking about the "+ C" and how it feels almost cosmetic. Here's how I make it feel important.
In differential equations class, we eventually get around to learning about general and particular solutions. Let's introduce the idea much earlier, like in early days of antiderivatives. So consider:
dy/dx = x
The complete solution to "y" will be the general solution ("y_g") plus the particular solution ("y_p").
To find the general solution, first solve dy/dx = 0. The solution to that is "y_g = C". You'll always get "y_g = C" if what's on the left is "dy/dx".
Now we find the particular solution to "dy/dx = x"; we get "y _p= x^2 / 2".
Add the particular and general solutions, and you get "y = x^2 / 2 + C". We got the complete answer, not by slapping an arbitrary "+ C" on there, but because it actually belongs.
Same approach would apply if the initial problem were "dy/dx + y = x". First we'd need to solve "dy/dx + y = 0", and it WOULDN'T be "y_g = C", but rather "y_g = C*e^(-x)". The particular solution would be "y_p = x - 1", leading to a total solution of "y = x - 1 + C*e^(-x)". Same methodology, very few arbitrary steps.
Why not using a hyperbolic substitution? In this case, x = sinh(t) and dx = cosh(t) dt 🙂 Then use cosh²(t) - sinh²(t) = 1, so the integrand simplifies to cosh²(t) (or -cosh²(t)) which is easily integrated to
t/2 + 1/2 sinh(t) cosh(t) + c or
1/2 arsinh(x) + 1/2 x sqrt(x² + 1) + c after back-substitution (and -t/2 - 1/2 sinh(t) cosh(t) + c or -1/2 arsinh(x) - 1/2 x sqrt(x² + 1) + c for the other case).
Learned to appreciate hyperbolic substitutions after reading this great book: Khristo Boyadzhiev, Special Techniques For Solving Integrals: Examples And Problems 🙂
Correct! Mathematica spits it out exactly the same way.
Addendum: with TrigToExp resp. ExpToTrig you can switch between these two forms.
Problems like these normally gives me trouble because I’m very likely to make a mistake somewhere when I’m using the first method.
I play with ODEs like
(1-x^2)y'' - xy' + n^2y=0
(1-x^2)y'' - 2xy' + n(n+1)y=0
and problem for me is not solving it with power series but with condition y(1)=1
I like the trig sub method myself. X = tan(theta). Dx would be (sec(theta))^2 d(theta).
dy/dx=√x²+1.....(Only taking positive square root)
dy=√x²+1 dx
Put x =tan∆ thus dx=sec²∆ d∆
dy=sec∆ sec²∆ d∆
By integration by parts,u=sec∆,v=sec²∆
y=1/2[sec∆ tan∆+ln|sec∆+tan∆|]+c
Resubstituting for x,
y=1/2[(√1+x²)x+ln(|√(x²+1)+x|)]
Does Sybermath offer a calc 1 course?
dy/dx = ax + b
You didn't use fact that ln(t) = -ln(1/t)
so your final answer using Euler's substitution is wrong
problem
Your two methods have different answers. How come there's a negative in front of the ln in the answer for method 1, but not in the answer for method 2?
(dy/dx)² - 1 = x²
(dy/dx)² = x² + 1
dy/dx = ±√(x² + 1)
∫ dy = ±∫ √(x² + 1) dx
y = ±∫ √(x² + 1) dx
Solution is for the positive branch and the negative will also be a solution. The integral will be completed and the constant of integration added to the right side at the end.
y = ∫ √(x² + 1) dx
Let
x = tan t
t = tan⁻¹ x
dx = sec² t dt
y = ∫ sec³ t dt
= ∫ (1 + tan²t) sec t dt
= ∫ sec t dt + ∫ tan²t sec t dt
= ∫ sec t dt + ∫ sin²t /(cos³ t ) dt
Let
r = cos t
√(1-r²) = sin t
-ain t dt = dr
-dr = sin t dt
Note that dr has one of the sin t terms of sin²t and the other sin t is in the √(1-r²).
y = ∫ sec t dt - ∫ √(1-r²)/(r³) dr
Integrate ∫ √(1-r²)/(r³) dr by parts.
dv = dr /(r³)
u = √(1-r²)
du = -r dr /√(1-r²)
v = -(½) 1/(r²)
v du = dr /[2 r√(1-r²) ]
uv. = -(½) √(1-r²)/(r²)
y = ∫ sec t dt - [ uv - ∫ v du ]
= ∫ sec t dt - [ -(½) √(1-r²)/(r²) -
∫ dr /[2 r√(1-r²) ]]
= ∫ sec t dt +
√(1-r²)/(2r²) +
∫ dr /[2 r√(1-r²) ]]
Replace r = cos t
y = ∫ sec t dt +√(1-r²)/(2r²) - (½) ∫ sec t dt
y = (½) ∫ sec t dt + (1/2) tan t sec t
We know ∫ sec t dt = ln | sec t + tan t |.
y = (1/2) ln | sec t + tan t |+ (1/2) tan t sec t
t = tan⁻¹ x
x = tan t
√(x²+1) = sec t
2y = ln (√(1+x²) + x)+
x√(1+x²)
y = (½) ln (√(1+x²) + x )+ (½) x√(1+x²) + c
, where c is a constant of integration.
answer
y = ± ½ [ln (√(1+x²) + x )
+ x√(1+x²) ] + c,
c ∈ ℝ
I don't understand differential equations. No one has ever explained the concept clearly. And this video didn't help at all. I completed a bachelor's degree in mathematics but lost grades due to these. The lecturers were also useless at explaining these.
hocam 1. yol daha güzel.
Garezin mi var kardeşim trigoyla 3 satır
@lightyagami5365 yok garezim trigonometri yolu da iyi ben 1. Yolu daha çok beğendim
Nice! Another method could be the sub x=i sin(t), maybe you could make a video abot it at aplusbi 😊💯💢💥💪💪