Adding the snippet to your page:
Theme => Edit HTML =>
<head>
<script type="text/javascript" async
src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.2/MathJax.js?config=TeX-MML-AM_CHTML">
</script>
</head>
For example:
(1)
∞∑i=0xii!=limn→∞n∏i=1(1+aix)=x∫−∞eydy=ex
(2)
x∈N⇒∃y∈N:y=x+1(3)
∀x∈R,∃y∈R:x+y=0∄y∈Q:y2=2
(4)When a≠0, there are two solutions to ax2+bx+c=0 and they are x=−b±√b2−4ac2a.Or you may love using Online LaTeX formula editor and browser-based math equation editor
http://hostmath.com/
Another source:How-do-you-use-LaTeX-mathematical-on-Quora?
Another example:
What is the solution of the differential equation
(dydx)2+4ydydxtanx+4y2sec2x=cos6x+4y2
Let see how the problem will be solved by Yebraw Mada, studies Mathematics and Physics at University of Bristol (2021)
What a horrific looking equation. This may take some time to solve…
After a while of pondering I’ve realized that this is actually quite a lot simpler than it looks. We just need to rearrange the fuck out of it. Here goes:
We start with
(dydx)2+4ydydxtanx+4y2sec2x=cos6x+4y2
Let’s subtract the 4y2, giving us
(dydx)2+4ydydxtanx+4y2sec2x−4y2=cos6x
Now factorizing
(dydx)2+4ydydxtanx+4y2(sec2x−1)=cos6x
Now remembering our trig identities this becomes
(dydx)2+4ydydxtanx+4y2tan2x=cos6x
The next step is to multiply by cos2x
cos2x(dydx)2+4ydydxsinxcosx+4y2sin2x=cos8x
It is not so obvious to spot the next step, but notice that the LHS factorizes perfectly.
(dydxcosx+2ysinx)2=(cos4x)2
Note that here I will just focus on the positive square root of each side. The next method will apply for the negative values too.
dydxcosx+2ysinx=cos4x
Dividing through by cosx
dydx+2ytanx=cos3x
Now we solve using an integrating factor. Here it is
e∫2tanxdx
e2lnsecx
sec2x
Now multiplying by this integrating factor we get
dydxsec2x+2ysec2xtanx=sec2xcos3x
ddx(ysec2x)=cosx
We eventually end up with
y=cos2x(sinx+C)
Phew. That was tiring. This solution can then be checked by substituting it back into the original differential equation.
Thanks for the A2A!
x∈N⇒∃y∈N:y=x+1(3)
∀x∈R,∃y∈R:x+y=0∄y∈Q:y2=2
(4)When a≠0, there are two solutions to ax2+bx+c=0 and they are x=−b±√b2−4ac2a.Or you may love using Online LaTeX formula editor and browser-based math equation editor
http://hostmath.com/
Another source:How-do-you-use-LaTeX-mathematical-on-Quora?
Another example:
What is the solution of the differential equation
(dydx)2+4ydydxtanx+4y2sec2x=cos6x+4y2
Let see how the problem will be solved by Yebraw Mada, studies Mathematics and Physics at University of Bristol (2021)
What a horrific looking equation. This may take some time to solve…
After a while of pondering I’ve realized that this is actually quite a lot simpler than it looks. We just need to rearrange the fuck out of it. Here goes:
We start with
(dydx)2+4ydydxtanx+4y2sec2x=cos6x+4y2
Let’s subtract the 4y2, giving us
(dydx)2+4ydydxtanx+4y2sec2x−4y2=cos6x
Now factorizing
(dydx)2+4ydydxtanx+4y2(sec2x−1)=cos6x
Now remembering our trig identities this becomes
(dydx)2+4ydydxtanx+4y2tan2x=cos6x
The next step is to multiply by cos2x
cos2x(dydx)2+4ydydxsinxcosx+4y2sin2x=cos8x
It is not so obvious to spot the next step, but notice that the LHS factorizes perfectly.
(dydxcosx+2ysinx)2=(cos4x)2
Note that here I will just focus on the positive square root of each side. The next method will apply for the negative values too.
dydxcosx+2ysinx=cos4x
Dividing through by cosx
dydx+2ytanx=cos3x
Now we solve using an integrating factor. Here it is
e∫2tanxdx
e2lnsecx
sec2x
Now multiplying by this integrating factor we get
dydxsec2x+2ysec2xtanx=sec2xcos3x
ddx(ysec2x)=cosx
We eventually end up with
y=cos2x(sinx+C)
Phew. That was tiring. This solution can then be checked by substituting it back into the original differential equation.
Thanks for the A2A!
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