Question

i will give medal..
Obtain the general solution of differential equation and reverse the process.

xcos^2ydx + tany dy =0
x^2yy' = e^y
y' = ysecx

Answer 1

are these three seperate problems?

Answer 2

yes..

Answer 3

$$
\Large{
x\cos^2y ~dx + \tan y ~dy =0
\\~\\ \frac{ x\cos^2y ~dx + \tan y ~dy}{dx} = \frac{0}{dx}
\\ x\cos^2y + \tan y ~\frac{dy}{dx} =0
\\ \tan y ~\frac{dy}{dx} =- x\cos^2y
\\~\\ \frac{\tan y}{\cos^2y } ~\frac{dy}{dx} =- x
\\~\\ \frac{\tan y}{\cos^2y }~ dy=- x ~dx
\\~\\ \int \frac{\tan y}{\cos^2y }~ dy=\int -x ~dx
}
$$

Answer 4

can you integrate that, you might want to simplify it first

Answer 5

i don't know how to integrate the firsrt variable..

Answer 6

$$
\Large{
x\cos^2y ~dx + \tan y ~dy =0
\\~\\ \frac{ x\cos^2y ~dx + \tan y ~dy}{dx} = \frac{0}{dx}
\\ x\cos^2y + \tan y ~\frac{dy}{dx} =0
\\ \tan y ~\frac{dy}{dx} =- x\cos^2y
\\~\\ \frac{\tan y}{\cos^2y } ~\frac{dy}{dx} =- x
\\~\\ \frac{\tan y}{\cos^2y }~ dy=- x ~dx
\\~\\ \int \frac{\tan y}{\cos^2y }~ dy=\int -x ~dx
\\~\\ \int \frac{\frac{\sin y}{\cos y}}{ \cos^2y }~ dy=\int -x ~dx
\\~\\ \int \frac{\ \sin y}{ \cos^3y }~ dy=\int -x ~dx
}
$$

Answer 7

you can do u-substitution at this point

Answer 8

what do you mean>>??

Answer 9

let u = cos y , du = -sin y dy

Answer 10

Does that look familiar?

Answer 11

yes..

Answer 12

$$
\Large{
x\cos^2y ~dx + \tan y ~dy =0
\\~\\ \frac{ x\cos^2y ~dx + \tan y ~dy}{dx} = \frac{0}{dx}
\\ x\cos^2y + \tan y ~\frac{dy}{dx} =0
\\ \tan y ~\frac{dy}{dx} =- x\cos^2y
\\~\\ \frac{\tan y}{\cos^2y } ~\frac{dy}{dx} =- x
\\~\\ \frac{\tan y}{\cos^2y }~ dy=- x ~dx
\\~\\ \int \frac{\tan y}{\cos^2y }~ dy=\int -x ~dx
\\~\\ \int \frac{\frac{\sin y}{\cos y}}{ \cos^2y }~ dy=\int -x ~dx
\\~\\ \int \frac{ \sin y}{ \cos^3y }~ dy=\int -x ~dx
\\~\\ u = \cos y , ~ du = -\sin y~ dy,
\\~\\\frac{-1}{-1} \int \frac{ \sin y}{ \cos^3y }~ dy=\int -x ~dx
\\~\\\frac{1}{-1} \int \frac{- \sin y ~dy}{ \cos^3y }=\int -x ~dx
\\~\\\frac{1}{-1} \int \frac{du}{ u^3 }=\int -x ~dx

}$$

Answer 13

note that -1/ -1 = 1 , so i did not change the integral . just prepared it for substitution

Answer 14

okay...then??

Answer 15

$$
\Large{
x\cos^2y ~dx + \tan y ~dy =0
\\~\\ \frac{ x\cos^2y ~dx + \tan y ~dy}{dx} = \frac{0}{dx}
\\ x\cos^2y + \tan y ~\frac{dy}{dx} =0
\\ \tan y ~\frac{dy}{dx} =- x\cos^2y
\\~\\ \frac{\tan y}{\cos^2y } ~\frac{dy}{dx} =- x
\\~\\ \frac{\tan y}{\cos^2y }~ dy=- x ~dx
\\~\\ \int \frac{\tan y}{\cos^2y }~ dy=\int -x ~dx
\\~\\ \int \frac{\frac{\sin y}{\cos y}}{ \cos^2y }~ dy=\int -x ~dx
\\~\\ \int \frac{ \sin y}{ \cos^3y }~ dy=\int -x ~dx
\\~\\ u = \cos y , ~ du = -\sin y~ dy,
\\~\\\frac{-1}{-1} \int \frac{ \sin y}{ \cos^3y }~ dy=\int -x ~dx
\\~\\\frac{1}{-1} \int \frac{- \sin y ~dy}{ \cos^3y }=\int -x ~dx
\\~\\\frac{1}{-1} \int \frac{du}{ u^3 }=\int -x ~dx
\\~\\\frac{1}{-1} \int u^{-3}~du=\int -x ~dx
\\~\\\frac{1}{-1}\cdot \frac{u^{-3+1}}{(-3+1)}= -\frac{x^2}{2}+C
\\~\\-1\cdot \frac{u^{-2}}{(-2)}= -\frac{x^2}{2}+C
\\~\\ \frac{(\cos y)^{-2}}{2}= -\frac{x^2}{2}+C
\\~\\ \frac{1}{2 \cos^2 y }= -\frac{x^2}{2}+C
}$$

Answer 16

That is the general solution. To reverse the process i think they mean, take the derivative

Answer 17

thats a good way to check that you integrated correctly, by the way

Answer 18

$$\Large{
\\~\\ \frac{(\cos y)^{-2}}{2}= -\frac{x^2}{2}+C
\\~\\ 2\cdot \frac{(\cos y)^{-2}}{2}= 2\cdot \left(-\frac{x^2}{2}+C \right)
\\~\\ (\cos y)^{-2}= -x^2+2C
\\~\\ \rm take~ implicit~ derivative ~ wrt ~ x
\\~\\ \frac{d}{dx} (\cos y)^{-2}= \frac{d}{dx} \left(-x^2+2C \right)
\\~\\ -2 (\cos y)^{-3}\cdot \frac{dy}{dx} = -2x + 0
}
$$

Answer 19

now that doesnt look like the original differential equation, so we have to look at our original differential equation. can we bridge these two expressions

Answer 20

by algebra and trig manipulation

Answer 21

$$
\Large{
\\~\\ \frac{(\cos y)^{-2}}{2}= -\frac{x^2}{2}+C
\\~\\ 2\cdot \frac{(\cos y)^{-2}}{2}= 2\cdot \left(-\frac{x^2}{2}+C \right)
\\~\\ (\cos y)^{-2}= -x^2+2C
\\~\\ \rm take~ implicit~ derivative ~ wrt ~ x
\\~\\ \frac{d}{dx} (\cos y)^{-2}= \frac{d}{dx} \left(-x^2+2C \right)
\\~\\ -2 (\cos y)^{-3}\cdot \frac{dy}{dx} = -2x + 0
\\~\\ (\cos y)^{-3}\cdot \frac{dy}{dx} = x
\\~\\ \frac{1}{(\cos y)^{3}}\cdot \frac{dy}{dx} = x
\\~\\ \frac{1}{\cos^2 y }\cdot \frac{1}{\cos y}\cdot \frac{dy}{dx} = x
\\~\\ \frac{1}{\cos y}\cdot \frac{dy}{dx} = x\cdot \cos^2 y
\\~\\ \frac{\sin y}{\sin y}\cdot \frac{1}{\cos y}\cdot \frac{dy}{dx} = x\cdot \cos^2 y


}$$

Answer 22

siny/siny??
whre it came from??

Answer 23

i introduced it, but it might not help

Answer 24

1-sin^2y = cos^2y??

Answer 25

oh i made a mistake when i took derivative

Answer 26

grrrr

Answer 27

$$
\Large{
\\~\\ \frac{(\cos y)^{-2}}{2}= -\frac{x^2}{2}+C
\\~\\ 2\cdot \frac{(\cos y)^{-2}}{2}= 2\cdot \left(-\frac{x^2}{2}+C \right)
\\~\\ (\cos y)^{-2}= -x^2+2C
\\~\\ \rm take~ implicit~ derivative ~ wrt ~ x
\\~\\ \frac{d}{dx} (\cos y)^{-2}= \frac{d}{dx} \left(-x^2+2C \right)
\\~\\ -2 (\cos y)^{-3}\cdot \color{red}{ (-\sin y )} \frac{dy}{dx} = -2x + 0



} $$

Answer 28

it's okY

Answer 29

COS^2y * COSy ? so that sin y/cos y will become tany...

Answer 30

am i right?

Answer 31

$$
\Large{
\\~\\ \frac{(\cos y)^{-2}}{2}= -\frac{x^2}{2}+C
\\~\\ 2\cdot \frac{(\cos y)^{-2}}{2}= 2\cdot \left(-\frac{x^2}{2}+C \right)
\\~\\ (\cos y)^{-2}= -x^2+2C
\\~\\ \rm take~ implicit~ derivative ~ wrt ~ x
\\~\\ \frac{d}{dx} (\cos y)^{-2}= \frac{d}{dx} \left(-x^2+2C \right)
\\~\\ -2 (\cos y)^{-3}\cdot \color{red}{ (-\sin y )} \frac{dy}{dx} = -2x + 0
\\~\\ 2 (\cos y)^{-3}\cdot \color{red}{ (\sin y )} \frac{dy}{dx} = -2x
\\~\\ (\cos y)^{-3}\cdot \color{red}{ (\sin y )} \frac{dy}{dx} = -x
\\~\\ \frac{1}{(\cos y)^{3}}\cdot \color{red}{ (\sin y )} \frac{dy}{dx} = -x
\\~\\ \frac{1}{(\cos y)^{2}}\cdot \frac{1}{\cos y } (\sin y ) \frac{dy}{dx} = -x
\\~\\ \frac{1}{(\cos y)^{2}}\cdot \frac{\sin y}{\cos y } \cdot \frac{dy}{dx} = -x
} $$

Answer 32

right, see if you can finish it ☺