Solve the differential equation:
$cos^2x\frac{dy}{dx}+y=tanx$
#differential equations
jolly star
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mrEinpest #ripDarpingertheGOAT
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Solve the differential equation:
$cos^2x\frac{dy}{dx}+y=tanx$
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jolly star
What i tried:
Divided both sides by cos^2(x) so that dy/dx is free of it and I get
dy/dx+ysec^x=tanxsec^2(x)
I tried to get this into this form:
dy/dx+yf(x)=g(x) so that I can use the integrating factor which is e^∫f(x)dx
Here the integrating factor is e^∫sec^2(x)dx= e^tanx
Which gives,
ye^tanx=∫tanxsec^2(x).e^tanxdx
How do I solve this integral ∫tanxsec^2(x).e^tanxdx?? using by part?
I just let secx=t so that dt=secxtanxdx
which means the inetgral becomes ∫t.e^(dt/t)dt
e^dt/t is confusing me.
jolly star
What i tried:
Divided both sides by cos^2(x) so that dy/dx is free of it and I g...
For integral use u sub tan x=u
Then by parts
jolly star
For integral use u sub tan x=u
Oh okay that is much better than the confusing secx sub
it wasn't just confusing it was wrong
∫u.e^(u)du is what we get. Thank you.
U should have spotted tan x and it's derivative sec^2 x
jolly star
U should have spotted tan x and it's derivative sec^2 x
Yes, you're right, however I went the other way as I could see a way to get udu, but the problem was with e^tanx.
Anyways, thank you.
+close
foggy plaza