Question

Find a solution to:
y' = 1 + ytan x
when y(0) = 1

Answer 1

?

Answer 2

\[y'-\tan(x) y=1\]

Answer 3

we want to multiply both sides by
\[v(x)=e^{\int\limits_{}^{}tanx dx}=e^{\int\limits_{}^{}\frac{\sin(x)}{\cos(x)} dx} \]
let \[u=\cos(x)=> du=-\sin(x) dx\]
\[v(x)=e^{\int\limits_{}^{}\frac{-1}{u} du}=e^{-\ln(u)}=e^{-\ln(\cos(x))}=\frac{1}{\cos(x)}\]

Answer 4

\[\frac{1}{\cos(x)}y'-\frac{1}{\cos(x)}\tan(x)y=\frac{1}{\cos(x)}\]
\[\sec(x)y'-\sec(x)\tan(x)y=\sec(x)\]
we can write
\[(\sec(x) y)'=\sec(x)\]

Answer 5

now integrate both sides

Answer 6

\[\sec(x) y=\int\limits_{}^{}\sec(x) \cdot \frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)} dx +C\]
\[\sec(x) y=\int\limits_{}^{}\frac{du}{u}+C=\sec(x) y=\ln|u|+C=\ln|\sec(x)+\tan(x)|+C\]

Answer 7

so we have
\[\sec(x) y=\ln|\sec(x)+\tan(x)|+C\]
now when x=0,y=1
\[\sec(0)(1)=\ln|\sec(0)+\tan(0)|+C\]

Answer 8

solve for C

Answer 9

\[(1)(1)=\ln|(1)+(0)|+C\]

Answer 10

\[1=0+C\]

Answer 11

\[C=1 => \sec(x) y=\ln|\sec(x)+\tan(x)|+1\]

Answer 12

\[y =(\ln|\sec(x)+\tan(x)|+1)/\sec(x)\]

Answer 13

Thanks

Answer 14

np

Answer 15

hmmm...

Answer 16

I believe the integrating factor should be \[\cos(x)\]

Answer 17

i disagree

Answer 18

\[y'-\tan(x) y=1\]
\[y'+(-\tan(x))y=1\]

\[u=e^{\int-tan(x)dx}\]
\[=\cos(x)\]

Answer 19

must be in standard form

Answer 20

oops i forgot about the negative

Answer 21

fine you are right

Answer 22

lol

Answer 23

:)

Answer 24

I try to be when I can

Answer 25

makes the problem easier ... that's a good thing

Answer 26

\[\cos(x) y'-\cos(x)\tan(x)y=\cos(x)\]
\[(\cos(x) y)'=\cos(x)\]

Answer 27

\[\cos(x) y=\sin(x)+C\]

Answer 28

tons easier!

Answer 29

yep

Answer 30

\[\cos(0)(1)=\sin(0)+C =>C=1\]

Answer 31

\[\cos(x) y=\sin(x)+1\]

Answer 32

Wow, I though those secant integrals seemed a bit unexpected! Relieved