Question

solve the bernolli equation!

Answer 1

\[dy/dx -(y/100)=0\]

Answer 2

i know they are seperable,but can we do the bernoulli way?

Answer 3

No i think it can't be solved that way, we won't be able to make any substitutions here

Answer 4

solve another
\[dy/dx- 2y=\sin2x\]

Answer 5

\[\frac{dy}{dx}-2y=\sin 2x\]
\[y'+p(x)y=q(x)\]
\[p(x)=-2\]
mulitply both sides by \(e^{\int -2 dx}=e^{-2x}\)
\[e^{-2x}\frac{dy}{dx}-2ye^{-2x}=\sin 2x e^{-2x}\]
this is
\[\frac{d}{dx} (e^{-2x}y)=\sin 2x e^{-2x}\]
now integrate both sides to solve the differential equation
\[e^{-2x}y=\int \sin 2x e^{-2x} dx +C\]
Can you solve now?

Answer 6

lol main thing was to integrate this e-2xsin2x

Answer 7

okay I'll do it

Answer 8

ash wait if we take ln bith sides , then integrating wud become easy huh

Answer 9

ash am i right?

Answer 10

ash plz do it im stuck in itegration

Answer 11

Sorrw @wasiqss I was afk

Answer 12

No you can't take log, it won't help.
We have
\[ I=\int \sin x e^{-2x} dx
\]
Now we'll use integration by parts

Answer 13

i took ln and im stuck at inegration of lnsin2x, rest is done

Answer 14

If you take log on the right hand side you'll have
\[\ln {\int e^{-2x} sin x dx}\]
you can't take log inside the intgral

Answer 15

so what to do now

Answer 16

we have
\[\int uv= u\int v -\int (u'\ \int v)\]
We have
\[ I=\int \sin 2x e^{-2x} dx
\]
u= sin 2x
\(v=e^{-2x}\)
\[I= \sin 2x \int e^{-2x} dx- \int (\frac {d}{dx} \sin 2x \int e^{-2x} dx)\]
We get
\[I= sin 2x \times \frac{e^{-2x} }{-2} -\int ( 2 \cos 2x \times \frac{e^{-2x} }{-2} )dx\]
we get
\[I=-\sin 2x \frac{e^{-2x} }{2}+\int ( \cos 2x \times e^{-2x}) dx\]

Answer 17

It's still left. Did you understand till here?

Answer 18

yehhh did tilll hea next

Answer 19

\[I=-\sin 2x \frac{e^{-2x} }{2}+\int ( \cos 2x \times e^{-2x}) dx\]
Now \(u=\cos 2x \) and \(v=e^{-2x}\)
\[I=-\sin 2x \frac{e^{-2x} }{2}+\cos 2x\int e^{-2x} dx-\int (\frac{d}{dx} \cos 2x \int e^{-2x} dx) dx \]
We get now
\[I=-\sin 2x \frac{e^{-2x} }{2}+ \cos 2x \frac{e^{-2x} }{-2}-\int (-2\sin 2x \times \frac{e^{-2x} }{-2}) dx\]
we get
\[I=-\sin 2x \frac{e^{-2x} }{2}-\cos 2x \frac{e^{-2x} }{2}-\int \sin 2x \times e^{-2x }dx\]
@wasiqss What's the integral term ? Do you remember what it is?

Answer 20

I'm waiting for your response?

Answer 21

im seeing

Answer 22

we did its integration in earlier part so we jus have to replace now i guess

Answer 23

@wasiqss1 this is what we were trying to integrate
\[I=\int sin 2x \times e^{-2x} dx\]

\[I=-\sin 2x \frac{e^{-2x} }{2}+ \cos 2x \frac{e^{-2x} }{-2}-\int (\sin 2x \times e^{-2x} ) dx\]
so

\[I=-\sin 2x \frac{e^{-2x} }{2}+ \cos 2x \frac{e^{-2x} }{-2}-I\]
we get
\[2I=-\sin 2x \frac{e^{-2x} }{2}- \cos 2x \frac{e^{-2x} }{2}\]
or
\[I=-\sin 2x \frac{e^{-2x} }{4}- \cos 2x \frac{e^{-2x} }{4}\]

Answer 24

thanks a tonnes ash , u rock :) finally u deserve a medal

Answer 25

jk lol

Answer 26

Glad to help:D